The volume of 17.6 moles of helium at normal air pressure and room temperature is approximately 0.416 m³.
To determine the volume (V) of 17.6 moles of helium, we can use the ideal gas law equation: p⋅V = nRT.
Given:
Number of moles (n) = 17.6 moles
Pressure (p) = 101,000 N/m²
Temperature (T) = 20°C
First, we need to convert the temperature from Celsius to Kelvin. The conversion can be done by adding 273.15 to the Celsius value:
T(K) = T(°C) + 273.15
Converting the temperature:
T(K) = 20°C + 273.15 = 293.15 K
Next, we substitute the values into the ideal gas law equation:
p⋅V = nRT
Plugging in the values:
101,000 N/m² ⋅ V = 17.6 moles ⋅ 8.314 KJ/K ⋅ 293.15 K
Now, we can solve for the volume (V) by rearranging the equation:
V = (17.6 moles ⋅ 8.314 KJ/K ⋅ 293.15 K) / 101,000 N/m²
Calculating the volume:
V ≈ 0.416 m³
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−33.0 cm is used to form an image of an arrow that is 14.8 cm away from the mirror. If the arrow is 2.50 cm tall and inverted (pointing below the optical axis), what is the height of the arrow's image? (Include the sign of the value in your answer.)
The height of the image of the arrow formed by the mirror is -5.57 cm. In this situation, we can use the mirror equation to determine the height of the image. The mirror equation is given by:
1/f = 1/di + 1/do,
where f is the focal length of the mirror, di is the distance of the image from the mirror, and do is the distance of the object from the mirror.
Given that di = -33.0 cm and do = 14.8 cm, we can rearrange the mirror equation to solve for the focal length:
1/f = 1/di + 1/do,
1/f = 1/-33.0 + 1/14.8,
1/f = -0.0303 + 0.0676,
1/f = 0.0373,
f = 26.8 cm.
Since the mirror forms a virtual image, the height of the image (hi) can be determined using the magnification equation:
hi/h₀ = -di/do,
where h₀ is the height of the object. Given that h₀ = 2.50 cm, we can substitute the values into the equation:
hi/2.50 = -(-33.0)/14.8,
hi/2.50 = 2.23,
hi = 2.50 * 2.23,
hi = 5.57 cm.
Since the image is inverted, the height of the image is -5.57 cm.
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Consider the following potential: Voi x≤0 V(x) = {-Vo; 0 < x a Assuming that the flux of particles are incident from the right, and the energy of the particles are 0 < E< Vo, find the amplitude of the reflected wave in the region > a
The amplitude of the reflected wave in the region x > a is given by Ar = -Ai √(1 - E / Vo) e^(-ik1a).
The given potential is a step potential of height -Vo at x ≤ 0, and 0 at 0 < x < a, and height 0 beyond x > a.
The probability current density J for a particle of energy E in a given region is given as J = (h / 2πi) [ψ*(dψ / dx) - (dψ* / dx) ψ]where ψ is the wave function and ψ* is its complex conjugate.
Using the probability current density expression, we can write down the transmission and reflection coefficients. The transmission coefficient T is the probability flux transmitted through the barrier, and the reflection coefficient R is the probability flux reflected from the barrier. The probability flux J is proportional to the square of the amplitude of the wave. Thus, we can write the transmission and reflection coefficients as:
T = |At|² / |Ai|² and R = |Ar|² / |Ai|²
where At is the amplitude of the transmitted wave, Ar is the amplitude of the reflected wave, and Ai is the amplitude of the incident wave.
Now, let's solve the problem at hand.
A particle of energy E is incident from the right, with an amplitude of Ai. The wave function for the particle in the region x ≤ 0 is given as:
ψ1(x) = Ae^(ik1x) + Be^(-ik1x), where k1 = √(2m(E + Vo)) / h and A and B are constants.
The wave function for the particle in the region 0 < x < a is given as:
ψ2(x) = Ce^(ik2x) + De^(-ik2x), where k2 = √(2mE) / h and C and D are constants.
The wave function for the particle in the region x > a is given as:
ψ3(x) = Ee^(ik3x), where k3 = √(2mE) / h and E is a constant.
Note that we have assumed that the potential is zero in the region x > a.
Using the boundary conditions at x = 0 and x = a, we can solve for the constants A, B, C, D, and E in terms of Ai as follows:
A = Ai / 2 + Ar / 2, B = Ai / 2 - Ar / 2, C = Ae^(ik1a) + Be^(-ik1a), D = Ae^(-ik1a) + Be^(ik1a), and E = Ce^(ik2a).
Now, we can calculate the reflection and transmission coefficients as:
R = |Ar|² / |Ai|² = |B - Ai / 2|² / |Ai|² = |Ai / 2 - (Ai / 2) e^(-2ik1a)|² / |Ai|² = |1/2 - 1/2 e^(-2ik1a)|² = sin²(k1a)T = |At|² / |Ai|² = |E|² / |Ai|² = |Ce^(ik2a)|² / |Ai|² = |C|² / |Ai|² = 1 - sin²(k1a)
Thus, we have derived the reflection and transmission coefficients in terms of the incident amplitude Ai and the energy E of the particle. For particles with energy 0 < E < Vo, we have sin(k1a) = √(1 - E / Vo) and cos(k1a) = √(E / Vo). The amplitude of the reflected wave in the region x > a is given by Ar = -Ai / 2 e^(-ik1a) (1 - e^(-2ik1a)).Thus, we have Ar = -Ai sin(k1a) e^(-ik1a).
Hence, the amplitude of the reflected wave in the region x > a is given by Ar = -Ai √(1 - E / Vo) e^(-ik1a).
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please write a full paraphrasing for the text below. thanks
The magnitude of each of the electric forces with which two point charges at rest interact is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them and has the direction of the line that joins them. . The force is repulsive if the charges are of the same sign, and attractive if they are of the opposite sign. Coulomb's law does comply with the principles of superposition since it determines the electric force of attraction or repulsion experienced by a point charge in the presence of another. The electrical forces between two charges can vary since in some the charges or the distance between them are doubled.
The text states Coulomb's law which expresses that the magnitude of electric forces between two point charges, which are stationary, is proportional to both charges' magnitudes and inversely proportional to the distance square between them.
If two point charges are in the same direction, they repel, and if they are in opposite directions, they attract.Coulomb's law follows the superposition concept, which calculates the repulsion or attraction electric force between a point charge in the presence of another point charge. Due to the doubled distance or charges, the electrical forces between two charges may differ.
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Determine the upward force that the biceps muscle exerts when a 75 Newton load is held in the hand when the arm is at 900 angles as shown. If the combined weight of the forearm and hand is assumed to be 35 Newton’s and acts at the center of gravity.
The total upward force exerted by the biceps muscle when holding the 75 Newton load in the hand at a 90-degree angle is 110 Newtons
To determine the upward force exerted by the biceps muscle when holding a 75 Newton load in the hand at a 90-degree angle, we need to consider the forces acting on the arm. The total force exerted by the biceps muscle can be calculated by summing the upward force required to counteract the load's weight and the weight of the forearm and hand. Given that the combined weight of the forearm and hand is 35 Newtons and acts at the center of gravity, the force required to counteract this weight is 35 Newtons in the downward direction. To maintain equilibrium, the biceps muscle must exert an equal and opposite force of 35 Newtons in the upward direction. Additionally, since the load in the hand weighs 75 Newtons, the biceps muscle needs to exert an additional 75 Newtons in the upward direction to counteract its weight. Therefore, the total upward force exerted by the biceps muscle when holding the 75 Newton load in the hand at a 90-degree angle is 110 Newtons.
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62. 56. When Sputnik I was launched by the U.S.S.R. in October 1957, American scientists wanted to know as much as possible about this new artificial satellite. If Sputnik orbited Earth once every 96 min, calculate its orbital velocity and altitude. (6.2)
The orbital velocity of Sputnik I is 7.91 x 10³ m/s and its altitude is 0.75 x 10⁶ m.
When Sputnik I was launched by the U.S.S.R. in October 1957, American scientists wanted to know as much as possible about this new artificial satellite.
If Sputnik orbited Earth once every 96 min, calculate its orbital velocity and altitude. (6.2)
The expression for the period of revolution of an artificial satellite of mass m around a celestial body of mass M is given by,
T = 2π √ (R³/GM)
where, T = Period of revolution
R = Distance of the artificial satellite from the center of the earth
G = Universal Gravitational constant
M = Mass of the earth
For Sputnik I,
Period of revolution, T = 96 minutes (convert it to seconds)
T = 96 * 60
= 5760 seconds
Universal Gravitational constant,
G = 6.67 x 10⁻¹¹ Nm²/kg²
Mass of the earth, M = 5.98 x 10²⁴ kg
The altitude of Sputnik I from the surface of the earth can be calculated as,
Altitude = R - R(earth)where,
R(earth) = radius of the earth
= 6.4 x 10⁶ m
Orbital velocity of Sputnik I
Orbital velocity of Sputnik I can be calculated as,
v = 2πR/T
Substitute the value of
T = 5760 seconds and solve for v,
v = 2πR/5760m/s
Calculate R, we have
T = 2π √ (R³/GM)5760
= 2π √ (R³/(6.67 x 10⁻¹¹ x 5.98 x 10²⁴))
Solve for R,
R = (GMT²/4π²)¹/³
= [(6.67 x 10⁻¹¹ x 5.98 x 10²⁴) x (5760)²/4π²]¹/³
= 7.15 x 10⁶ m
Therefore,
Altitude = R - R(earth)
= 7.15 x 10⁶ m - 6.4 x 10⁶ m
= 0.75 x 10⁶ m
Orbital velocity, v = 2πR/T
= (2 x 3.14 x 7.15 x 10⁶ m)/5760 sec
= 7.91 x 10³ m/s
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It is found that a 122 cm long tube that is open on both ends resonates at frequencies of 700 Hz and 840 Hz, but not at any frequencies between these two. What is the speed of sound in the air in and around this tube? v = Number Units
The speed of sound in the air in and around the tube is 343 m/s.
The fundamental frequency of an open-ended tube is given by the following equation:
f = v / (2L)
where:
f is the fundamental frequency in hertz
v is the speed of sound in meters per second
L is the length of the tube in meters
In this case, the fundamental frequency is 700 Hz and the length of the tube is 122 cm. Plugging these values into the equation, we get the following speed of sound:
v = f * 2L = 700 Hz * 2 * 0.122 m = 343 m/s
The speed of sound in air is typically around 340 m/s, so this is a reasonable value.
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An ice skater begins a spin with her arms out. Her angular velocity at the beginning of the spin is 3.0 rad/s and his moment of inertia is 10.0 kgm 2 . As the spin proceeds she pulls in her arms, decreasing her moment of inertia to 8.0 kgm 2 . It takes her half a second to pull in her arms and change speeds.
a. What is her angular momentum before pulling in her arms?
b. What is her angular momentum after pulling in her arms?
c. What is her angular velocity after pulling in her arms?
d) Calculate α during the 0.5 seconds that she is extending her arms.
Any help is appreciated. Thank you in advance :)
a) Angular momentum before pulling in her arms: 30.0 kgm^2/s.
b) Angular momentum after pulling in her arms: 30.0 kgm^2/s.
c) Angular velocity after pulling in her arms: 3.75 rad/s.
d) Angular acceleration during arm extension: -7.5 rad/s^2.
To solve this problem, we can use the conservation of angular momentum, which states that the total angular momentum of a system remains constant unless acted upon by an external torque
a) Before pulling in her arms, her moment of inertia is 10.0 kgm^2 and her angular velocity is 3.0 rad/s.
The formula for angular momentum is L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Therefore, her angular momentum before pulling in her arms is L1 = (10.0 kgm^2)(3.0 rad/s) = 30.0 kgm^2/s.
b) After pulling in her arms, her moment of inertia decreases to 8.0 kgm^2.
The angular momentum is conserved, so the angular momentum after pulling in her arms is equal to the angular momentum before pulling in her arms.
Let's denote this angular momentum as L2.
L2 = L1 = 30.0 kgm^2/s.
c) We can rearrange the formula for angular momentum to solve for the angular velocity.
L = Iω -> ω = L/I.
After pulling in her arms, her moment of inertia is 8.0 kgm^2. Substituting the values, we get:
ω = L2/I = 30.0 kgm^2/s / 8.0 kgm^2 = 3.75 rad/s.
Therefore, her angular velocity after pulling in her arms is 3.75 rad/s.
d) To calculate the angular acceleration (α) during the 0.5 seconds while she is extending her arms, we can use the formula α = (ω2 - ω1) / Δt, where ω2 is the final angular velocity, ω1 is the initial angular velocity, and Δt is the time interval.
Since she is extending her arms, her moment of inertia increases back to 10.0 kgm^2.
We know that her initial angular velocity is 3.75 rad/s (from part c).
Δt = 0.5 s.
Plugging in the values, we get:
α = (0 - 3.75 rad/s) / 0.5 s = -7.5 rad/s^2.
The negative sign indicates that her angular acceleration is in the opposite direction of her initial angular velocity.
To summarize:
a) Angular momentum before pulling in her arms: 30.0 kgm^2/s.
b) Angular momentum after pulling in her arms: 30.0 kgm^2/s.
c) Angular velocity after pulling in her arms: 3.75 rad/s.
d) Angular acceleration during arm extension: -7.5 rad/s^2.
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The deep end of a pool is 2.67 meters. What is the water pressure at the bottom of the deep end? Density of water: 1000 kg/m3
The water pressure at the bottom of the deep end of the pool is 26,370 Pascals (Pa).
To calculate the water pressure, we can use the formula:
Pressure = Density × Gravity × Height
Density of water = 1000 kg/m^3
Height = 2.67 meters
Gravity = 9.8 m/s^2 (approximate value)
Plugging in the values:
Pressure = 1000 kg/m^3 × 9.8 m/s^2 × 2.67 meters
Pressure ≈ 26,370 Pa
Therefore, the water pressure at the bottom of the deep end of the pool is approximately 26,370 Pascals.
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This table shows Wayne’s weight on four different planets.
Planet Wayne’s weight
(pounds)
Mars 53
Neptune 159
Venus 128
Jupiter 333
Arrange the planets in decreasing order of their strength of gravity.
Answer: Jupiter > Neptune > Venus > Mars
Explanation: edmentum
A copper calorimetric cup with a mass of 100g contains 96g of water at 13 C. If 70g of a substance at 84 degC is dropped into the cup, the temperature increases to 20 degC. Find the specific heat capacity of the substance.
Someone pours 150g of heated lead shot into a 250g aluminum calorimeter cup that contains 200g of water at 25 degC . The final temperature is 28 degC. What was the intial temperature of the lead shot?
What mass of water at 50 degC can be converted into steam at 110 degC by 9.6 x10^6 J?
Answer: The mass of water required is 4247.79 g (answer).
Therefore, the mass of water at 50°C that can be converted into steam at 110°C by 9.6 × 106 J is 4247.79 g.
Question 1 : A copper calorimetric cup with a mass of 100g contains 96g of water at 13 C. If 70g of a substance at 84 degC is dropped into the cup, the temperature increases to 20 degC. Find the specific heat capacity of the substance.
Solution :The amount of heat lost by hot body = amount of heat gained by cold body
Applying the formula of specific heat capacity
mcΔT = msΔT
Since there is no loss of heat to the surrounding mcΔT = msΔT
m1c1ΔT1 = m2s2ΔT2
where m1, c1 and ΔT1 are the mass, specific heat capacity and the temperature change of the copper cup and water.
m2, s2 and ΔT2 are the mass, specific heat capacity and the temperature change of the substance.
We know that the mass of copper calorimetric cup = 100g
the mass of water = 96g
the temperature of water = 13°C
the mass of the substance = 70g
the temperature of the substance = 84°C
The final temperature after mixing = 20°C
Temperature change of the substance,
ΔT2 = Final temperature - initial temperature
= 20°C - 84°C= - 64°C
Temperature change of the water,
ΔT1 = Final temperature - initial temperature
= 20°C - 13°C= 7°C
Thus, by substituting the values in the formula:
m1c1ΔT1 = m2s2ΔT2(100 g) (0.385 J/g°C) (7°C)
= (70 g) s2 (-64°C)s2
= 0.448 J/g°C
Specific heat capacity of the substance is 0.448 J/g°C (answer)
Hence, the specific heat capacity of the substance is 0.448 J/g°C.
Question 2: Someone pours 150g of heated lead shot into a 250g aluminum calorimeter cup that contains 200g of water at 25°C. The final temperature is 28°C. What was the initial temperature of the lead shot?
Solution:
Heat lost by lead shot = Heat gained by water + Heat gained by Aluminium container Q1 = Q2 + Q3
The formula of heat: Q = m × c × ΔT
Where,Q1 = Heat lost by lead shot
m = mass of the object
c = Specific heat capacity
ΔT = Temperature difference.
Q2 = Heat gained by water
m = mass of the object
c = Specific heat capacity
ΔT = Temperature difference.
Q3 = Heat gained by Aluminium container
m = mass of the object
c = Specific heat capacity
ΔT = Temperature difference.
Substitute the values given in the question,Q1 = (150 g) × c × (Ti - 28) °C
Q2 = (200 g) × 4.18 J/g°C × (28 - 25) °C
= 2502 JQ3 = (250 g) × 0.897 J/g°C × (28 - 25) °C
= 672.75 J Q1 = Q2 + Q3(150 g) × c × (Ti - 28) °C
= 2502 J + 672.75 J(150 g) × c × (Ti - 28) °C
= 3174.75 J(150 g) × c × (Ti - 28) / 150 g
= 3174.75 J / 150 gTi - 28
= 21.16°C (Approx.)Ti
= 49.16°C (answer)
Hence, the initial temperature of the lead shot was 49.16°C.
Question 3 : What mass of water at 50°C can be converted into steam at 110°C by 9.6 x 10^6 J?
Solution:
To find the mass of water, we use the formula, Q = mL
Where,
Q = Amount of heat required to change the phase of water from liquid to gas
L = Latent heat of vaporisation
m = Mass of water required.
To find the value of L, we use the specific heat capacity of water.The amount of heat required to raise 1 g of water by 1°C = 1 cal/g°C
Specific heat capacity of water = 4.18 J/g°C
Amount of heat required to raise 1 g of water by 1°C = 4.18 J/g°C
Specific latent heat of vaporisation of water = 2260 J/g
Amount of heat required to convert 1 g of water into steam = 2260 J/g
To find the mass of water,m = Q / LWhere,
Q = 9.6 × 106 J (Given)
L = 2260 J/g
Substitute the given values in the formula,
m = 9.6 × 106 J / 2260 J/g
m = 4247.79 g (Approx.)
Hence, the mass of water required is 4247.79 g (answer).
Therefore, the mass of water at 50°C that can be converted into steam at 110°C by 9.6 × 106 J is 4247.79 g.
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Compare a 1kg solid gold bar or a 15g solid gold wedding ring, which has a higher (i) density (ii) specific gravity? (i) bar, (i) bar
(i) ring, (ii) ring
(i) same, (ii) same
(i) bar, (ii) ring
(i) bar, (ii) same
(i) ring, (ii) bar
(i) ring, (ii) same
(i) same, (ii) bar
(i) same, (ii) ring
Please document your reasoning
A 1kg solid gold bar or a 15g solid gold wedding ring, which has a higher (i) The density of the gold bar and gold ring is the same.
(ii) The specific gravity of the gold bar and gold ring is the same.
(i) Density:
Density is defined as the mass of an object divided by its volume. The density of a substance remains constant regardless of the size or shape of the object. In this case, we are comparing a 1 kg solid gold bar and a 15 g solid gold wedding ring.
Given:
Mass of gold bar = 1 kg
Mass of gold ring = 15 g
Since density is calculated by dividing mass by volume, we need to consider the volume of the objects as well. The volume of an object is directly proportional to its mass.
Assuming that both the gold bar and gold ring are made of the same material (gold) with the same density, the density of gold will be the same for both objects. Therefore, the answer is (i) same.
(ii) Specific Gravity:
Specific gravity is the ratio of the density of a substance to the density of a reference substance. The reference substance is usually water at a standard temperature and pressure. Since we are comparing two gold objects, the reference substance will remain the same.
The specific gravity of gold is typically measured with respect to water. The density of gold is much higher than that of water, so the specific gravity of gold is greater than 1.
Again, assuming that both the gold bar and gold ring are made of the same material (gold), their specific gravities will be the same as the specific gravity is determined by the density of the substance relative to water. Therefore, the answer is (ii) same.
In summary:
(i) The density of the gold bar and gold ring is the same.
(ii) The specific gravity of the gold bar and gold ring is the same.
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X-rays with an energy of 339 keV undergo Compton scattering with a target. If the scattered X-rays are detected at 57.7^{\circ}∘relative to the incident X-rays, what is the wavelength of the scattered photon?
Answer:
The
wavelength
of the scattered photon is approximately 1.11 × 10^(-11) meters.
Explanation:
Compton scattering is a phenomenon where X-rays interact with electrons, resulting in a shift in wavelength. To determine the wavelength of the scattered photon, we can use the Compton scattering formula:
Δλ = λ' - λ = λ_c * (1 - cos(θ))
Where:
Δλ is the change in wavelength
λ' is the wavelength of the scattered photon
λ is the wavelength of the incident X-ray photon
λ_c is the Compton wavelength (approximately 2.43 × 10^(-12) m)
θ is the scattering angle
Given:
Energy of the incident X-ray photon (E) = 339 keV = 339 * 10^3 eV
Scattering angle (θ) = 57.7 degrees
First, let's calculate the wavelength of the incident X-ray photon using the energy-wavelength relationship:
E = hc / λ
Where:
h is Planck's constant (approximately 6.63 × 10^(-34) J·s)
c is the speed of light (approximately 3.00 × 10^8 m/s)
Converting the energy to joules:
E = 339 * 10^3 eV * (1.60 × 10^(-19) J/eV) = 5.424 × 10^(-14) J
Rearranging the equation to solve for λ:
λ = hc / E
Substituting the values:
λ = (6.63 × 10^(-34) J·s * 3.00 × 10^8 m/s) / (5.424 × 10^(-14) J) ≈ 1.22 × 10^(-11) m
Now, let's calculate the change in wavelength using the Compton scattering formula:
Δλ = λ_c * (1 - cos(θ))
Substituting the values:
Δλ = (2.43 × 10^(-12) m) * (1 - cos(57.7 degrees))
Calculating cos(57.7 degrees):
cos(57.7 degrees) ≈ 0.551
Δλ = (2.43 × 10^(-12) m) * (1 - 0.551) ≈ 1.09 × 10^(-12) m
Finally, we can calculate the wavelength of the scattered photon by subtracting the change in wavelength from the wavelength of the incident X-ray photon:
λ' = λ - Δλ
Substituting the values:
λ' = (1.22 × 10^(-11) m) - (1.09 × 10^(-12) m) ≈ 1.11 × 10^(-11) m
Therefore, the wavelength of the scattered photon is approximately 1.11 × 10^(-11) meters.
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The voltage and Power ratings of a Particular light bulb, which are It's normal operating values are lov and 60w. Assume the resistance of the of with ating Conditions. If the light bulb is operated with a Current that is 50% of the current. rating Idrawn by the bulb? of the bulb, what is the actual Power
The actual power consumed is 30W when the light bulb is worked with a current that is 50% of its current rating using Ohm's Law.
Normal operating value = 60W
Bulb operation = 50% of current.
The relation between voltage, current, and resistance is given by Ohm's Law.
V = I * R.
R = V / I
The formula used for calculating the power rating in normal operating conditions is:
P_0 = V_0 * I_0
The actual current drawn by the bulb I_actual is:
V_0 = I_actual * R
R = V_0 / I_actual
P_actual = V_0 * I_actual
Substituting the values we get:
P_actual = V_0 * I_actual = V_0 * (0.5 * I_0)
60W = V_0 * I_0
V_0 = 60W / I_0
P_actual = (60W / I_0) * (0.5 * I_0) = 30W
Therefore, we can conclude that the actual power consumed is 30W.
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(a) the energy released per event in joules ] (b) the change in mass (in kg ) during the event ×kg [0/1.92 Points] SERCP11 30.4.OP.021. In a pair-production reaction, a photon produces a muon-antimuon pair. γ→μ −
+μ +
The rest energy of a muon is 105.7MeV. (a) What is the lowest possible frequency (in Hz ) of the photon that can produce the muon-antimuon pair? Hz (b) What is the wavelength (in m ) that corresponds to this lowest possible frequency? 2s What is the relationship between frequency, wavelength, and the speed of light? m
Lowest possible frequency: 4.84 x 10^20 Hz, Corresponding wavelength: 6.19 x 10^-13 m (or 2s), The relationship between frequency, wavelength, and the speed of light is given by c = fλ.
The lowest possible frequency (f) of the photon that can produce the muon-antimuon pair can be found by using the equation E = hf, where E is the energy (rest energy of the muon in this case) and h is the Planck's constant (approximately 6.63 x 10^-34 J·s). Converting the rest energy of the muon from MeV to joules (1 MeV = 1.6 x 10^-13 J), we have E = 105.7 MeV = 105.7 x 1.6 x 10^-13 J. By rearranging the equation, we can solve for the frequency: f = E / h. Plugging in the values, we get f = (105.7 x 1.6 x 10^-13 J) / (6.63 x 10^-34 J·s) ≈ 4.84 x 10^20 Hz. (b) The relationship between frequency (f), wavelength (λ), and the speed of light (c) is given by the equation c = fλ, where c is the speed of light (approximately 3 x 10^8 m/s). Rearranging the equation, we can solve for the wavelength: λ = c / f. Plugging in the values, we get λ = (3 x 10^8 m/s) / (4.84 x 10^20 Hz) ≈ 6.19 x 10^-13 m or 2s (as mentioned in the question).
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An RL circuit is composed of a 12 V battery, a 6.0 H inductor and a 0.050 Ohm resistor. The switch is closed at t=0 The time constant is 2.0 minutes and after the switch has been closed a long time the voltage across the inductor is 12 V. The time constant is 1.2 minutes and after the switch has been closed a long time the voltage across the inductor is zero. The time constant is 2.0 minutes and after the switch has been closed a long time the voltage across the inductor is zero
The time constant is 1.2 minutes and after the switch has been closed a long time the voltage across the inductor is 12 V.
The RL circuit described has a time constant of 1.2 minutes, and after the switch has been closed for a long time, the voltage across the inductor is 12 V.
The time constant (τ) of an RL circuit is determined by the product of the resistance (R) and the inductance (L) and is given by the formula τ = L/R. In this case, the time constant is 1.2 minutes.
When the switch is closed, current begins to flow through the circuit. As time progresses, the current increases and approaches its maximum value, which is determined by the battery voltage and the circuit's total resistance.
In an RL circuit, the voltage across the inductor (V_L) can be calculated using the formula V_L = V_0 * (1 - e^(-t/τ)), where V_0 is the initial voltage across the inductor, t is the time, and e is the base of the natural logarithm.
Given that the voltage across the inductor after a long time is 12 V, we can set V_L equal to 12 V and solve for t to determine the time it takes for the voltage to reach this value. The equation becomes 12 = 12 * (1 - e^(-t/τ)).
By solving this equation, we find that t is equal to approximately 3.57 minutes. Therefore, after the switch has been closed for a long time, the voltage across the inductor in this RL circuit reaches 12 V after approximately 3.57 minutes.
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Five points per problem. 1. A spring is used to launch a 200 g dart horizontally off of a 5 m tall building. The spring has constant k=120 N/m and was compressed 0.04 m. How far in the horizontal direction from where it was shot does the dart land, if it falls a total of 5 m ? Recall the spring potential energy is given by SPE =0.5 ∗k∗ x∧ 2. 2. A bicycle wheel with moment of inertia 1=0.2kgm ∧
2 is accelerated from rest to 30 rad/s in 0.4 s. If the force of the chain is exerted 0.1 m from the pivot, what is the magnitude of the force? 3. A 30 kg dog jumps from rest and reaches a maximum height of 2 m. What is the net force acting on the dog in the upward direction if it acts for 0.8s while he is jumping? 4. A hanging 3 kg. im long fluorescent light is supported on each end by a wire. If the weight of the lamp is evenly distributed, what is the tension in each wire? 5. Two kids are sitting on either side of the pivot of a 15 kg.2 m long seesaw. The pivot is displaced by 0.3 m away from the center of mass of the seesaw. Each child is sitting at the end of the board. If one child is 30 kg. and the seesaw is perfectly balanced, what is the mass of the other child? 6. A cube of ice (literally a cube, with side length 0.02 m and density 0.92 kg/m ∧
3 ) is floating in vodka (density 0.95 kg/m ∧
3 ). What is the fraction of the ice submerged in the vodka if it is in equilibrium?
The answer is 1.
1. Given data: Mass of dart, m = 200 g = 0.2 kg,
Height of building, h = 5 m, Spring constant,
k = 120 N/m, Distance of compression, x = 0.04 m,
Total distance fallen, y = 5 m.
The spring potential energy is given by the relation, SPE = 0.5 * k * x²
The spring potential energy is equal to the kinetic energy of the dart when the spring is released.
Let v be the velocity with which the dart is launched.
The kinetic energy of the dart is given by, KE = (1/2) * m * v²
Applying conservation of energy between potential energy and kinetic energy,
SPE = KE0.5 * k * x²
= (1/2) * m * v²
= sqrt( k * x² / m )Given that the total distance fallen by the dart is y = 5 m and that it was launched horizontally, the time taken for it to reach the ground is given by,
t = sqrt( 2 * y / g )
where g is the acceleration due to gravity.
Using the time taken and the horizontal velocity v, we can determine the horizontal distance traveled by the dart as follows,
Distance = v * t = sqrt( 2 * k * x² * y / (g * m) )
The required distance is Distance = sqrt( 2 * 120 * 0.04² * 5 / (9.81 * 0.2) ) = 1.13 m.
2. Given data: Moment of inertia, I = 0.2 kg m²,
Angular velocity, ω = 30 rad/s,
Time taken, t = 0.4 s,
Distance from pivot, r = 0.1 m.
The torque exerted on the wheel is given by,
T = Iαwhere α is the angular acceleration.
The angular acceleration is given by,α = ω / t The force F applied by the chain causes a torque about the pivot given by,τ = Fr
The magnitude of the force F is then given by,F = τ / r
Substituting the values, I = 0.2 kg m², ω = 30 rad/s,
t = 0.4 s, r = 0.1 m,
we getα = ω / t = 75 rad/s²τ
= Fr = IαF
= τ / r = Iα / r
= (Iω / t) / r
= (0.2 * 30 / 0.4) / 0.1
= 15 N
3. Given data: Mass of dog, m = 30 kg, Maximum height reached, h = 2 m, Time taken, t = 0.8 s.
The net force acting on the dog in the upward direction while it is jumping is given by the relation,
F = mgh / t
where g is the acceleration due to gravity.
Substituting the values, m = 30 kg,
h = 2 m,
t = 0.8 s,
g = 9.81 m/s²,
we get F = mg h / t = (30 * 9.81 * 2) / 0.8
= 735.75 N
4. Given data: Mass of lamp, m = 3 kg, Length of lamp, L = 1 m.
The weight of the lamp acts vertically downwards. The two wires exert equal and opposite tensions T on the lamp, at angles of θ with the vertical.
Resolving the tensions into horizontal and vertical components, Tsin(θ) = mg / 2and,
Tcos(θ) = T cos (θ)We have two equations and two unknowns (T and θ).
Dividing the two equations above, Tsin (θ) / T cos(θ) = (mg / 2) / T cos(θ)tan(θ)
= mg / 2Tcos(θ)²
= T² - Tsin²(θ)
= T² - (mg / 2)²
Substituting the values, m = 3 kg,
L = 1 m, g = 9.81 m/s², we get tan(θ) = 3 * 9.81 / 2 = 14.715
T cos(θ)² = T² - (3 * 9.81 / 2)²
Solving for T cos (θ) and T sin(θ),T cos(θ) = 11.401 N
T sin(θ) = 7.357 N
The tension in each wire is T = √(Tcos (θ)² + Tsin (θ)²) = 13.601 N
5. Given data: Mass of seesaw, m = 15 kg, Length of seesaw, L = 2 m,
Distance of pivot from center of mass, d = 0.3 m, Mass of one child, m1 = 30 kg, Mass of other child, m2 = ?
The seesaw is in equilibrium and hence the net torque about the pivot is zero. The net torque about the pivot is given by,
τ = (m1g)(L/2 - d) - (m2g)(L/2 + d)
where g is the acceleration due to gravity. Since the seesaw is in equilibrium, the net force acting on it is zero and hence we have,
F = m1g + m2g = 0
Substituting m1 = 30 kg,
L = 2 m, d = 0.3 m,
we get,τ = (30 * 9.81)(1.7) - (m2 * 9.81)(2.3) = 0
Solving for m2, we get m2 = (30 * 9.81 * 1.7) / (9.81 * 2.3) = 19.23 kg.
6. Given data: Density of ice, ρi = 0.92 kg/m³, Side length of cube, s = 0.02 m, Density of vodka, ρv = 0.95 kg/m³.
Let V be the volume of the ice cube that is submerged in the vodka. The volume of the ice cube is s³ and the volume of the displaced vodka is also s³.
Since the ice cube is floating, the weight of the displaced vodka is equal to the weight of the ice cube. The weight of the ice cube is given by, Wi = mgi
where gi is the acceleration due to gravity and is equal to 9.81 m/s².
The weight of the displaced vodka is given by, Wv = mvdg where dg is the acceleration due to gravity in vodka.
We have, dg = g (ρi / ρv)The fraction of the ice cube submerged in the vodka is given by,V / s³ = Wv / Wi
Substituting the values, gi = 9.81 m/s², dg = 9.81 * (0.92 / 0.95),
we get V / s³ = Wv / Wi
= (ρv / ρi) * (dg / gi)
= (0.95 / 0.92) * (0.92 / 0.95)
= 1.
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A ray of light traveling in water is incident on an interface with a flat piece of glass. The wavelength of the light in the water is 722 nm and its wavelength in the glass is 543 nm. If the ray in water makes an angle of 45.0 ∘
with respect to the normal to the interface, what angle does the refracted ray in the glass make with respect to the normal?
The refracted ray in the glass makes an angle of approximately 48.4° with respect to the normal.
To determine the angle of the refracted ray in the glass, we can use Snell's Law, which relates the angles and indices of refraction of light as it passes through different mediums. Snell's Law states that the ratio of the sines of the angles of incidence (θ₁) and refraction (θ₂) is equal to the ratio of the indices of refraction (n₁ and n₂) of the two mediums.
In this case, the incident angle in water (θ₁) is given as 45.0°, the wavelength of light in water (λ₁) is 722 nm, and the wavelength of light in glass (λ₂) is 543 nm.
We know that the index of refraction (n) of a medium is inversely proportional to the wavelength of light passing through it, so we can use the ratio of the wavelengths to calculate the ratio of the indices of refraction:
n₁ / n₂ = λ₂ / λ₁
Substituting the given values, we have:
n₁ / n₂ = 543 nm / 722 nm
To simplify the calculation, we can convert the wavelengths to meters:
n₁ / n₂ = (543 nm / 1) / (722 nm / 1) = 0.751
Now, we can apply Snell's Law:
sin(θ₁) / sin(θ₂) = n₂ / n₁
sin(θ₂) = (n₁ / n₂) * sin(θ₁)
Plugging in the values, we get:
sin(θ₂) = 0.751 * sin(45.0°)
To find the angle θ₂, we can take the inverse sine (or arcsine) of both sides:
θ₂ = arcsin(0.751 * sin(45.0°))
Evaluating this expression, we find:
θ₂ ≈ 48.4°
Therefore, the refracted ray in the glass makes an angle of approximately 48.4° with respect to the normal.
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A sculpture weighing 35000 N rests on a horizontal surface at the top of a 1.8 m high stand (Figure 2). The stand's cross-sectional area is 7.3 x 102 m2 and it is made of granite with a
Young's modulus of 4.5 x 1010 Pa. By how much does the sculpture compress the stand?
[3]
Figure 2
A. 1.9 x 10-2 mm
B. 5.2 x 102 mm
C. 32.85 x 10-2 mm
D. 6.3 x 102 mm
The sculpture compresses the stand by correct option A) 1.9 x 10-2 mm. Compression can be determined by dividing the applied force by the product of the cross-sectional area and the material's Young's modulus.
To calculate the compression of the stand, we can use Hooke's Law, which states that the deformation of a material is directly proportional to the applied force and inversely proportional to its stiffness or Young's modulus.
The weight of the sculpture is 35000 N, and it applies a force on the stand. This force causes the stand to compress.
Using the formula for compression, Δx = F/(A * E), where Δx is the compression, F is the force, A is the cross-sectional area, and E is the Young's modulus of the material, we can calculate the compression of the stand.
Δx = (35000 N) / ((7.3 x [tex]10^{2}[/tex] [tex]m^{2}[/tex]) * (4.5 x [tex]10^{10}[/tex] Pa))
Simplifying the expression, we find that the sculpture compresses the stand by approximately 1.9 x [tex]10^{-2}[/tex] mm.
Therefore, the correct answer is A. 1.9 x 10-2 mm.
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A fighter plane flying at constant speed 450 m/s and constant altitude 1000 m makes a turn of curvature radius 4000 m. On the ground, the plane's pilot weighs (61 kg)(9.8 m/s2 )=597.8 N. What is his/her apparent weight during the plane's turn? Answer in units of N.
The pilot's apparent weight during the plane's turn is 3665.3 N.
To determine the apparent weight of the pilot during the plane's turn, we need to consider the centripetal force acting on the pilot due to the turn. The apparent weight is the sum of the actual weight and the centripetal force.
Calculate the centripetal force:
The centripetal force (Fc) can be calculated using the equation[tex]Fc = (m * v^2) / r[/tex], where m is the mass of the pilot, v is the velocity of the plane, and r is the radius of curvature.
Fc = [tex](61 kg) * (450 m/s)^2 / 4000 m[/tex]
Fc = 3067.5 N
Calculate the apparent weight:
The apparent weight (Wa) is the sum of the actual weight (W) and the centripetal force (Fc).
Wa = W + Fc
Wa = 597.8 N + 3067.5 N
Wa = 3665.3 N
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When you go out to your car one cold winter morning you discover a 0.50-cm thick layer of ice on the windshield, which has an area of 1.9 m? . If the temperature of the ice is -3.0
°C, and its density is 917 kg/m', find the heat required to melt all the ice
The heat required to melt the 0.50-cm thick layer of ice on the 1.9 m² windshield is approximately 2,958,319.3 J.
To calculate the heat required to melt all the ice, we need to consider the energy required for both raising the temperature of the ice to its melting point and then melting it.
First, let's calculate the mass of the ice. The volume of the ice can be determined using its thickness and the area of the windshield:
Volume = Thickness * Area = (0.50 cm * 1.9 m²) = 0.0095 m³
Next, we can calculate the mass of the ice using its density:
Mass = Density * Volume = (917 kg/m³ * 0.0095 m³) = 8.71 kg
To raise the temperature of the ice from -3.0°C to its melting point (0°C), we need to provide energy using the specific heat capacity of ice. The specific heat capacity of ice is approximately 2.09 J/g°C.
First, let's convert the mass of ice to grams:
Mass (grams) = Mass (kg) * 1000 = 8.71 kg * 1000 = 8710 g
The energy required to raise the temperature of the ice can be calculated using the formula:
Energy = Mass * Specific Heat Capacity * Temperature Change
Energy = 8710 g * 2.09 J/g°C * (0°C - (-3.0°C)) = 8710 g * 2.09 J/g°C * 3.0°C = 49,179.3 J
Next, we need to consider the energy required to melt the ice. The latent heat of fusion for ice is approximately 334,000 J/kg.
The total energy required to melt the ice can be calculated as:
Energy = Mass * Latent Heat of Fusion
Energy = 8.71 kg * 334,000 J/kg = 2,909,140 J
Finally, we can calculate the total heat required to melt all the ice by adding the energy required for raising the temperature and melting the ice:
Total Heat = Energy for Temperature Change + Energy for Melting
Total Heat = 49,179.3 J + 2,909,140 J = 2,958,319.3 J
Therefore, the heat required to melt all the ice is approximately 2,958,319.3 J.
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A 230 g , 25-cm-diameter plastic disk is spun on an axle through its center by an electric motor.
What torque must the motor supply to take the disk from 0 to 1500 rpm in 5.0 s ? Express your answer in newton-meters.
To calculate the torque required to accelerate the plastic disk, calculate the moment of inertia (I) using the given mass and diameter. Then, calculate the initial angular velocity (ω0) by dividing the final angular velocity by the time. Using the change in rotational kinetic energy (ΔKE) and the change in angle (Δθ), the torque (τ) can be determined by dividing ΔKE by 2π.
To calculate the torque required to accelerate the plastic disk from 0 to 1500 rpm in 5.0 seconds, we need to use the rotational kinetic energy formula:
Rotational Kinetic Energy (KE) = (1/2) * Moment of Inertia * Angular Velocity^2
The moment of inertia (I) for a solid disk rotating about its central axis is given by:
Moment of Inertia (I) = (1/2) * Mass * Radius^2
Mass of the plastic disk (m) = 230 g = 0.23 kg
Diameter of the disk (d) = 25 cm = 0.25 m
Time (t) = 5.0 s
Final angular velocity (ω) = 1500 rpm = 1500 * (2π/60) rad/s (converting rpm to rad/s)
First, we need to calculate the moment of inertia (I) using the given mass and diameter:
I = (1/2) * m * (r^2)
= (1/2) * 0.23 kg * (0.125 m)^2
= 0.002875 kg·m^2
Next, we can calculate the initial angular velocity (ω0) by dividing the final angular velocity (ω) by the time (t):
Initial angular velocity (ω0) = ω / t
= (0 rad/s - 1500 * (2π/60) rad/s) / 5.0 s
= -1500 * (2π/60) / 5.0 rad/s
Now, we can calculate the change in rotational kinetic energy (ΔKE) by subtracting the initial rotational kinetic energy from the final rotational kinetic energy:
ΔKE = KE - KE0
= (1/2) * I * ω^2 - (1/2) * I * ω0^2
Finally, the torque (τ) required can be calculated using the equation:
ΔKE = τ * Δθ
where Δθ is the change in angle (2π radians).
Since we are going from 0 to a final angular velocity, Δθ is equal to 2π radians. Substituting the values into the equation, we can solve for the torque (τ).
ΔKE = τ * Δθ
τ = ΔKE / Δθ
τ = ΔKE / (2π)
Calculating this expression will give us the torque required in newton-meters.
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6. GO A plate carries a charge of 3.0 uC, while a rod carries a charge of +2.0 uC. How many electrons must be transferred from the plate to the rod, so that both objects have the same charge?
Approximately 6.24 x 10¹² electrons must be transferred from the plate to the rod for both objects to have the same charge.
To determine the number of electrons that must be transferred from the plate to the rod, we need to consider the elementary charge and the difference in charge between the two objects.
The elementary charge is the charge carried by a single electron, which is approximately 1.602 x 10⁻¹⁹ coulombs (C). The charge carried by an electron is approximately -1.602 x 10⁻¹⁹ coulombs (C).
Given that the plate carries a charge of 3.0 μC (microcoulombs) and the rod carries a charge of +2.0 μC, we need to find the difference in charge between them.
Converting the charges to coulombs:
Plate charge = 3.0 μC = 3.0 x 10⁻⁶ C
Rod charge = +2.0 μC = 2.0 x 10⁻⁶ C
The difference in charge is:
Difference in charge = Plate charge - Rod charge
= 3.0 x 10⁻⁶ C - 2.0 x 10⁻⁶ C
= 1.0 x 10⁻⁶ C
Since the plate has an excess of charge, electrons need to be transferred to the rod, which has a positive charge. The charge of an electron is -1.602 x 10^-19 C, so the number of electrons transferred can be calculated as:
Number of electrons transferred = Difference in charge / Charge of an electron
= 1.0 x 10⁻⁶ C / (1.602 x 10⁻¹⁹ C)
≈ 6.24 x 10¹² electrons
Therefore, approximately 6.24 x 10¹² electrons must be transferred from the plate to the rod for both objects to have the same charge.
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The monthly (30 days) electric bill included the cost of running a central air-conditioning unit for 2.5 hr/day at 4500 w, and a series connection of ten 4 W light bulbs for 7.5 hr/day. According to the energy company's recent tariff, electricity costs 2.06 TL per kWh. a) How much did these items contribute to the cost of the monthly electric bill? TL b) What if you were using 60 w light bulbs? TL
We need to determine the energy consumed by each appliance and then multiply it by the electricity cost per kilowatt-hour (kWh). The cost can be calculated using the power consumption and the duration of use for each appliance.
a) To calculate the cost contributed by the central air-conditioning unit, we first convert the power consumption from watts to kilowatts by dividing it by 1000. Then, we multiply the power consumption (4.5 kW) by the daily usage time (2.5 hours) and the number of days in a month (30) to obtain the energy consumption in kilowatt-hours. Finally, we multiply the energy consumption by the electricity cost per kWh (2.06 TL) to determine the cost contributed by the air-conditioning unit.
To calculate the cost contributed by the series connection of light bulbs, we calculate the total power consumption by multiplying the power consumption of each bulb (4 W) by the number of bulbs (10). Then, we multiply the total power consumption (40 W) by the daily usage time (7.5 hours) and the number of days in a month (30) to obtain the energy consumption in kilowatt-hours. Finally, we multiply the energy consumption by the electricity cost per kWh (2.06 TL) to determine the cost contributed by the light bulbs.
b) If we were using 60 W light bulbs instead of 4 W bulbs, we would repeat the calculations by replacing the power consumption of each bulb with 60 W. This would result in a higher total power consumption for the light bulbs, leading to a higher cost contributed by the light bulbs on the monthly electric bill.
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In a Photoelectric effect experiment, the incident photons each has an energy of 4.713×10 −19 J. The power of the incident light is 0.9 W. (power = energy/time) The work function of metal surface used is W 0 = 2.71eV. 1 electron volt (eV)=1.6×10 −19 J. If needed, use h=6.626×10 −34 J⋅s for Planck's constant and c=3.00×10 8 m/s for the speed of light in a vacuum. Part A - How many photons in the incident light hit the metal surface in 7.0 s ? Part B - What is the max kinetic energy of the photoelectrons? Part C - Use classical physics fomula for kinetic energy, calculate the maximum speed of the photoelectrons. The mass of an electron is 9.11×10 −31 kg
The incident photons energy is 1.337 × 10²². The max kinetic energy of the photoelectrons is 6.938 × 10⁻¹ eV. The maximum speed of the photoelectrons is 5.47 × 10⁵ m/s. The correct answer for a) 1.337 × 10²² photons b) 6.938 × 10⁻¹ eV c) 5.47 × 10⁵ m/s
Part A The power of the incident light, P = 0.9 W Total energy delivered, E = P x tE = 0.9 x 7 = 6.3 JThe energy of each photon, E = 4.713 × 10⁻¹⁹ J Number of photons, n = E/E = 6.3/4.713 × 10⁻¹⁹ = 1.337 × 10²² photons
Part B The energy of a photon = hν, where ν is the frequencyν = c/λ where c = speed of light and λ is the wavelength of light.λ = hc/E = hc/ (4.713 × 10⁻¹⁹) = 1.324 × 10⁻⁷ m Kinetic energy of a photoelectron is given by KE max = hν - W₀ = hc/λ - W₀ = (6.626 × 10⁻³⁴ × 3.0 × 10⁸)/1.324 × 10⁻⁷ - (2.71 × 1.6 × 10⁻¹⁹) = 1.11 × 10⁻¹⁹ J = 6.938 × 10⁻¹ eV
Part C Maximum speed of a photoelectron can be calculated by using classical mechanics equation: KEmax = (1/2)mv²where m is the mass of electron and v is the maximum speed. Rearranging gives: v = √(2KEmax/m) = √(2(6.938 × 10⁻¹ eV)(1.6 × 10⁻¹⁹ J/eV)/(9.11 × 10⁻³¹ kg)) = 5.47 × 10⁵ m/s (to 3 significant figures) Answer:a) 1.337 × 10²² photonsb) 6.938 × 10⁻¹ eVc) 5.47 × 10⁵ m/s
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Suppose that not all but only 50% of the neutrons were consumed in Big Bang Nucleosynthesis. What would the H:He mass ratio be?
The H:He mass ratio if only 50% of neutrons were used in Big Bang Nucleosynthesis will be 3:1.
Let us see how this conclusion was reached.
Big Bang Nucleosynthesis is a cosmological event in which the nuclei of helium, lithium, and deuterium were formed within a few seconds of the Big Bang. This event happened between 10 seconds and 20 minutes after the Big Bang and produced the elements that make up the universe. It is important to note that in this process, only some of the neutrons present were used. This is because most of the neutrons decayed into protons. This means that only about one neutron out of every seven was available to make heavier nuclei.
Suppose 7 neutrons were present during Big Bang Nucleosynthesis, and only 50% of them were used. Therefore, only 3.5 neutrons would have been used in the process. If we rounded that to 3 neutrons, the remaining neutrons would have decayed to form protons. This means that 6 protons and 3 neutrons would have combined to form helium-3 (2 protons and 1 neutron) and helium-4 (2 protons and 2 neutrons).
The H:He mass ratio would be calculated as follows:
For H, we have 2 protons, which is equivalent to a mass number of 2.
For He, we have 2 protons and 2 neutrons, which is equivalent to a mass number of 4.
Therefore, the H:He mass ratio is: 2:4, which is equivalent to 1:2, which can be further simplified to 3:1. Hence, the H:He mass ratio if only 50% of neutrons were used in Big Bang Nucleosynthesis would be 3:1.
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8) If the refracting index of light in a medium is n = 2.7, what is the speed of light in the medium? Find the wavelength of an EM wave with a frequency of 12 x 10° Hz in the medium with n = 2.7.
The speed of light in the medium with a refractive index of 2.7 is approximately 1.11 x 10⁸ meters per second. The wavelength of the EM wave is approximately 9.25 meters.
The speed of light in a medium can be calculated using the formula v = c/n, where v is the speed of light in the medium, c is the speed of light in a vacuum, and n is the refractive index of the medium.
In this case, the refractive index of the medium is given as n = 2.7. The speed of light in a vacuum is approximately 3 x 10⁸ meters per second.
Plugging these values into the formula, we get
v = (3 x 10⁸ m/s) / 2.7. Simplifying this expression gives us v ≈ 1.11 x 10^8 meters per second.
Therefore, the speed of light in the medium with a refractive index of 2.7 is approximately 1.11 x 10⁸ meters per second.
To find the wavelength of an electromagnetic wave with a frequency of 12 x 10⁶ Hz in the medium with n = 2.7, we can use the formula λ = v/f, where λ is the wavelength, v is the speed of light in the medium, and f is the frequency of the wave.
Using the previously calculated speed of light in the medium (v = 1.11 x 10⁸ m/s) and the given frequency (f = 12 x 10⁶ Hz), we can calculate the wavelength:
λ = (1.11 x 10⁸ m/s) / (12 x 10⁶ Hz) ≈ 9.25 meters.
Therefore, the wavelength of the EM wave with a frequency of 12 x 10⁶ Hz in the medium with n = 2.7 is approximately 9.25 meters.
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QUESTION 7 At an orange juice plant, orange juice pulp with a density of 1.25 g/mi passes through a pumping station where it is raised vertically by 575m at the rate 11,040.000 as per day. The liquid enters and leaves the pumping station at the same speed and through pass of opaal diameter. Determine the outpu mechanical power (in W) of the sit station fgnore any energy loss due to friction QUESTION An estimated force-time curve for a baseball struck by a bot is shown in the figure (file in Course Content) Let max 16,000 N. 15 ms, and th-2 ms. From this curve, determine the average force (in kN) exerted on the bal QUESTION 9 A billiard ball moving at 5.20 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.41 m/s at an angle of respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's speed after the collision QUESTION 10 3 points 5 points S points
As the liquid enters and leaves the pumping station at the same speed, it means that there is no net work done, and the output mechanical power of the sit station is zero (0).
The maximum force (Fmax) is 16,000 N, time is 15 ms, and t1/2 is 2 ms.From the graph, we can calculate the average force exerted on the baseball using the formula;Favg
= [tex]∆p/∆t[/tex]where ∆p
= mv - mu is the change in momentum, which can be calculated using the formula; ∆p
= m(v-u)
= F∆t, where F is the force and ∆t is the time.Favg
= [tex]F∆t/∆t[/tex]
= FThe average force exerted on the baseball is equal to the maximum force, Favg
= Fmax
= 16,000 N.Question 9:
The billiard ball moving at 5.20 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.41 m/s at an angle of θ
= 37° to the original line of motion. Conservation of momentum and kinetic energy can be applied to solve this problem.Before the collision, the momentum of the system is given as;p
= mu + 0
= muAfter the collision, the momentum of the system is given as;p'
= m1v1' + m2v2'where v1' and v2' are the final velocities of the two balls, and m1 and m2 are the masses of the two balls.Using the conservation of momentum, we can equate these two expressions;p
= p'mu
= [tex]m1v1' + m2v2'... (1)[/tex]
Kinetic energy is also conserved in elastic collisions.
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The velocity field of a flow is given by v= 6xi+ 6yj-7 tk.
a) Determine the velocity at a point x= 10 m; y = 6m; when t = 10 sec. Draw, approximately, a set of streamlines for the flow at instant t = 0.
b) Determine the acceleration field of the flow and the acceleration of the particle at the point and instant specified above. at the point and instant specified above
" The velocity at the point (x = 10 m, y = 6 m, t = 10 s) is 60i + 36j - 70k m/s.The acceleration of the particle at the point (x = 10 m, y = 6 m, t = 10 s) is -7k m/s²." Acceleration is a fundamental concept in physics that measures the rate of change of velocity of an object over time. It is defined as the derivative of velocity with respect to time.
a) To determine the velocity at the specified point (x = 10 m, y = 6 m, t = 10 s), we substitute these values into the given velocity field equation:
v = 6xi + 6yj - 7tk
v = 6(10)i + 6(6)j - 7(10)k
= 60i + 36j - 70k
Therefore, the velocity at the point (x = 10 m, y = 6 m, t = 10 s) is 60i + 36j - 70k m/s.
b) The acceleration field (a) can be obtained by taking the time derivative of the velocity field:
a = dv/dt = d(6xi + 6yj - 7tk)/dt
= 6(dxi/dt) + 6(dyj/dt) - 7(dtk/dt)
= 6(0i) + 6(0j) - 7k
= -7k
Therefore, the acceleration field is a = -7k m/s².
To determine the acceleration of the particle at the specified point (x = 10 m, y = 6 m, t = 10 s), we substitute these values into the acceleration field equation:
a = -7k
a = -7(1)k
= -7k
So, the acceleration of the particle at the point (x = 10 m, y = 6 m, t = 10 s) is -7k m/s².
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(40%) A standard device for measuring viscosities is the cone-and-plate viscometer, as shown in the figure below. A pool of liquid is placed on a flat stationary plate, which is brought into contact with an inverted cone. Torque measurements are made with the top piece, of radius R, rotated at an angular velocity while the bottom piece stationary. The angle ß between the surface of the cone and plate is small. Spherical coordinates (r, 0, 4) are used in the analysis, such that the rotation is in + direction and the cone and plate surfaces in contact with the fluid are given by 0=ande =-B, respectively. a) Show that a velocity field of the form V = V(r, 0) and V₁ = V₂ = 0 is consistent with differential mass conservation; b) The measurements are performed in the viscous flow regime when inertial terms in flow equations are negligible. What is the corresponding condition in terms of the problem parameters? c) Assuming that Stokes' equations are applicable, show that V = rf (0) is consistent with conservation of momentum. Do this by deriving the differential equation and boundary conditions for f(0) (do not solve this equation!); d) Instead of solving the equation derived in (c) in spherical coordinates, for << 1 it is possible to approximate the solution by the flow between two parallel plates in Cartesian coordinates. In such case the local height of the fluid between the plates is b = r sin ß-rß. Show that the approximate solution is of the form: wr V₂ = (1-0) B e) Using the result in (d) find the torque exerted on the bottom plate (at 0 = π/2) by the liquid from: T₂ = - Splate "ToodA, where top is the relevant component of the viscous stress tensor in spherical coordinates and dA = rdrdp. B R ZA liquid
A) A velocity field of the form V = V(r, θ) and V₁ = V₂ = 0 is consistent with differential mass conservation.
B) The condition for the measurements to be performed in the viscous flow regime, where inertial terms in flow equations are negligible, is when the Reynolds number (Re) is small. The Reynolds number is given by Re = (ρVd) / μ, where ρ is the density of the fluid, V is the characteristic velocity, d is the characteristic length scale, and μ is the dynamic viscosity of the fluid. When Re << 1, the inertial terms can be neglected.
C) Assuming Stokes' equations are applicable, a velocity field of the form V = r∇f(θ) is consistent with conservation of momentum. By deriving the differential equation and boundary conditions for f(θ), we can show this.
D) When β << 1, an approximation can be made by considering the flow between two parallel plates in Cartesian coordinates. In this case, the local height of the fluid between the plates is given by b = r sin β - rβ. The approximate solution for the velocity field in this configuration is of the form V₂ = (1 - cos β) β.
Using the result from the approximation in (D), we can find the torque exerted on the bottom plate at θ = π/2 by the liquid. The torque (T₂) is given by
[tex]T_2 = -\int\limits {dx S_plate (τ_top)dA} \,[/tex]
Where τ_top is the relevant component of the viscous stress tensor in spherical coordinates and dA = rdrdθ.
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Part A A curve of radius 71 m is banked for a design speed of 95 km/h. If the coefficient of static friction is 0.30 (wet pavement), at what range of speeds can a car safely make the curve? (Hint: Consider the direction of the friction force when the car goes too slow or too fast.] Express your answers using two significant figures separated by a comma. Vo ΑΣΦ o ? Omin, Omax = km/h Submit Request Answer
The car can safely make the curve within a speed range of approximately 59 km/h to 176 km/h considering the coefficient of static friction of 0.30 and a curve radius of 71 m.
The key concept to consider is that the friction force between the car's tires and the road surface provides the centripetal force required to keep the car moving in a curved path. The friction force acts inward and is determined by the coefficient of static friction (μs) and the normal force (N).
When the car goes too slow, the friction force alone cannot provide enough centripetal force, and the car tends to slip outward. In this case, the gravitational force component perpendicular to the surface provides the remaining centripetal force.
The maximum speed at which the car can safely make the curve occurs when the friction force reaches its maximum value, given by the equation:μsN = m * g * cos(θ),where m is the mass of the car, g is the acceleration due to gravity, and θ is the angle of banking. Rearranging the equation, we can solve for the normal force N:N = m * g * cos(θ) / μs.
The maximum speed (Omax) occurs when the friction force is at its maximum, which is equal to the static friction coefficient multiplied by the normal force:Omax = sqrt(μs * g * cos(θ) * r).Substituting the given values into the equation, we get:Omax = sqrt(0.30 * 9.8 * cos(θ) * 71).Similarly, when the car goes too fast, the friction force is not necessary to provide the centripetal force, and it tends to slip inward.
The minimum speed at which the car can safely make the curve occurs when the friction force reaches its minimum value, which is zero. This happens when the car is on the verge of losing contact with the road surface. The minimum speed (Omin) can be calculated using the equation: Omin = sqrt(g * tan(θ) * r).
Substituting the given values, we get:Omin = sqrt(9.8 * tan(θ) * 71).Therefore, the car can safely make the curve within a speed range of approximately 59 km/h to 176 km/h (rounded to two significant figures), considering the coefficient of static friction of 0.30 and a curve radius of 71 m.
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