The final velocity is approximately 1.74272 × 10¹ m/s.
To find the final velocity, we can use the kinematic equation:
v = u + at,
where
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time.
Given:
Initial velocity (u) = + 3.5200 × 10 m/s
Acceleration (a) = 5.68 × 10 m/s²
Time (t) = 2.440 × 10 seconds
Substituting these values into the equation, we have:
v = 3.5200 × 10 m/s + 5.68 × 10 m/s² × 2.440 × 10 seconds.
v = (3.5200 + 5.68 × 2.440) × 10 m/s.
v = (3.5200 + 13.9072) × 10 m/s.
v = 17.4272 × 10 m/s.
v = 1.74272 × 10¹ m/s.
Therefore, the final velocity is approximately 1.74272 × 10¹ m/s.
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Calculate the force of attraction between an electron and a proton located 2.0 mm apart.
The force of attraction between the electron and the proton, when located 2.0 mm apart, is approximately -2.304 x 10⁻⁸ N.
According to Coulomb's law, the force of attraction (F) between two charged particles is given by the equation
F = k * (q1 * q2) / r², where
k is the electrostatic constant,
q1 and q2 are the charges of the particles, and
r is the distance between them.
In this case, we have an electron with charge q1 = -1.6 x 10⁻¹⁹ C and a proton with charge q2 = +1.6 x 10⁻¹⁹ C. The distance between them is given as r = 2.0 mm, which is equivalent to 2.0 x 10⁻³ m.
The electrostatic constant, k, has a value of approximately 9.0 x 10⁹ Nm²/C².
Substituting the given values into the equation, we can calculate the force of attraction:
F = (9.0 x 10⁹ Nm²/C²) * ((-1.6 x 10⁻¹⁹ C) * (1.6 x 10⁻¹⁹ C)) / (2.0 x 10⁻³ m)²
Performing the calculations:
F ≈ -2.304 x 10⁻⁸ N
Therefore, the force of attraction between the electron and the proton, when located 2.0 mm apart, is approximately -2.304 x 10⁻⁸ N. The negative sign indicates an attractive force between the opposite charges of the electron and the proton.
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Suppose that you built the circuit given in Figure 3-2 of your lab instructions with slide wire of total length 7.4cm and try to experimentally determine the value of the unknown resistance Rx where Rc is 7.2. If the point of balance of the Wheatstone bridge you built is reached when 12 is 3.6 cm, calculate the experimental value for Rx. Give your answer in units of Ohms with i decimal. Answer:
To calculate the experimental value for Rx, we can use the concept of the Wheatstone bridge. In a balanced Wheatstone bridge, the ratio of resistances on one side of the bridge is equal to the ratio on the other side. The experimental value for Rx is approximately 3.79 Ω.
In this case, we have Rc = 7.2 Ω and the slide wire of total length is 7.4 cm. The point of balance is reached when 12 is at 3.6 cm.
To find the experimental value of Rx, we can use the formula:
Rx = (Rc * Lc) / Lx
Where Rx is the unknown resistance, Rc is the known resistance, Lc is the length of the known resistance, and Lx is the length of the unknown resistance.
Substituting the values into the formula:
Rx = (7.2 Ω * 3.6 cm) / (7.4 cm - 3.6 cm)
Rx ≈ 14.4 Ω / 3.8 cm
Rx ≈ 3.79 Ω
Therefore, the experimental value for Rx is approximately 3.79 Ω.
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Rope A has 2 times the length, 3 times the mass, and is under 5 times the tension that rope B is under. If transverse waves travel on both ropes, what is the ratio of the speed of the Wave on rope A to the speed of the wave on rope B ?
The ratio of the speed of the wave on rope A to the speed of the wave on rope B is 1.29.
According to the given statement, rope A is longer, heavier and under higher tension than rope B. As a result, the speed of waves in rope A will be greater than the speed of waves in rope B.
And the ratio of the speed of the wave on rope A to the speed of the wave on rope B can be determined by using the following formula's ∝ √(Tension/ mass) When everything else is held constant, the speed of a wave on a string is directly proportional to the square root of the tension on the string and inversely proportional to the square root of the linear density of the string.
So, the speed of the waves in rope A, VA can be written as
:vA = k√(TA/MA) ------ equation 1And the speed of waves in rope B, VB can be written as:
vB = k√(TB/MB) ------ equation 2Where k is a constant of proportionality that is constant for both equations.
Dividing equation 1 by equation 2 we get, VA/vB = √(TA/MA) / √(TB/MB)Taking the given information, we have:
Rope A has twice the length of Rope B, i.e., L_A=2L_BRope A has three times the mass of Rope B, i.e., M_A=3M_BRope A is under 5 times the tension of Rope B, i.e., T_A=5T_B
Replacing the values in equation we get, vA/vB = √(TA/MA) / √(TB/MB)= √ (5T_B / 3M_B) / √(T_B / M_B)= √(5/3)= 1.29
Therefore, the ratio of the speed of the wave on rope A to the speed of the wave on rope B is 1.29.
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An entity is in a 2-D infinite well of dimension 0≤x≤a 0 ≤ y ≤ b The wave function of this entity is given by y(x, y) = C sin(kxx) sin(k¸y) (a) Determine the values of kx, ky, and C.
For an entity in a 2D infinite well, the values of kx and ky are determined by the boundary conditions kx * 0 = 0 and ky = mπ / b, where m is a positive integer and the value of C is determined by normalizing the wave function through the integral ∫∫ |C sin(kx x) sin(ky y)|² dxdy = 1.
In a 2-D infinite well of dimensions 0 ≤ x ≤ a and 0 ≤ y ≤ b, the wave function for the entity is given by:
y(x, y) = C sin(kx x) sin(ky y)
To determine the values of kx, ky, and C, we need to apply the boundary conditions for the wave function.
Boundary condition along the x-direction:
Since the well extends from 0 to a in the x-direction, the wave function should be zero at both x = 0 and x = a. Therefore, we have:
y(0, y) = 0
y(a, y) = 0
Using the given wave function, we can substitute these values and solve for kx.
0 = C sin(kx * 0) sin(ky y)
0 = C sin(kx a) sin(ky y)
Since sin(0) = 0, we get:
kx * 0 = nπ
kx a = mπ
Here, n and m are positive integers representing the number of nodes along the x-direction and y-direction, respectively.
Since kx * 0 = 0, we have n = 0 (the ground state) for the x-direction.
For the y-direction, we have ky = mπ / b.
Normalization condition:
The wave function should also be normalized, which means the integral of the absolute square of the wave function over the entire 2-D well should be equal to 1.
∫∫ |y(x, y)|² dxdy = 1
∫∫ |C sin(kx x) sin(ky y)|² dxdy = 1
Using the properties of sine squared, the integral simplifies to:
C² ∫∫ sin²(kx x) sin²(ky y) dxdy = 1
Integrating over the ranges 0 to a for x and 0 to b for y, we can evaluate the integral.
Once we have the integral, we can set it equal to 1 and solve for C to determine its value.
Thus, the boundary conditions kx * 0 = 0 and ky = mπ / b are used to determine the values of kx and ky and the value of C is determined by normalizing the wave function.
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The gravitational force changes with altitude. Find the change in gravitational force for someone who weighs 760 N at sea level as compared to the force measured when on an airplane 1600 m above sea level. You can ignore Earth's rotation for this problem. Use a negative answer to indicate a decrease in force.
For reference, Earth's mean radius (RE) is 6.37 x 106 m and Earth's mass (ME) is 5.972 x 1024 kg. [Hint: take the derivative of the expression for the force of gravity with respect to r, such that Aweight dF dr Ar. Evaluate the derivative at
Substituting the given values for Earth's mean radius (RE) and Earth's mass (ME), as well as the weight of the individual[tex](m1 = 760 N / 9.8 m/s^2 = 77.55 kg)[/tex], we can calculate the change in gravitational force.
To find the change in gravitational force experienced by an individual weighing 760 N at sea level compared to the force measured when on an airplane 1600 m above sea level, we can use the equation for gravitational force:
[tex]F = G * (m1 * m2) / r^2[/tex]
Where:
F is the gravitational force,
G is the gravitational constant,
and r is the distance between the centers of the two objects.
Let's denote the force at sea level as [tex]F_1[/tex] and the force at 1600 m above sea level as [tex]F_2[/tex]. The change in gravitational force (ΔF) can be calculated as:
ΔF =[tex]F_2 - F_1[/tex]
First, let's calculate [tex]F_1[/tex] at sea level. The distance between the individual and the center of the Earth ([tex]r_1[/tex]) is the sum of the Earth's radius (RE) and the altitude at sea level ([tex]h_1[/tex] = 0 m).
[tex]r_1 = RE + h_1 = 6.37 * 10^6 m + 0 m = 6.37 * 10^6 m[/tex]
Now we can calculate [tex]F_1[/tex] using the gravitational force equation:
[tex]F_1 = G * (m_1 * m_2) / r_1^2[/tex]
Next, let's calculate [tex]F_2[/tex] at 1600 m above sea level. The distance between the individual and the center of the Earth ([tex]r_2[/tex]) is the sum of the Earth's radius (RE) and the altitude at 1600 m ([tex]h_2[/tex] = 1600 m).
[tex]r_2[/tex] = [tex]RE + h_2 = 6.37 * 10^6 m + 1600 m = 6.37 * 10^6 m + 1.6 * 10^3 m = 6.3716 * 10^6 m[/tex]
Now we can calculate [tex]F_2[/tex] using the gravitational force equation:
[tex]F_2[/tex] = G * ([tex]m_1 * m_2[/tex]) /[tex]r_2^2[/tex]
Finally, we can find the change in gravitational force by subtracting [tex]F_1[/tex] from [tex]F_2[/tex]:
ΔF = [tex]F_2 - F_1[/tex]
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The gravitational force acting on the person has decreased by 0.104 N when they are on an airplane 1600 m above sea level as compared to the force measured at sea level.
Gravitational force is given by F = G (Mm / r²), where G is the universal gravitational constant, M is the mass of the planet, m is the mass of the object, and r is the distance between the center of mass of the planet and the center of mass of the object.Given,At sea level, a person weighs 760N.
On an airplane 1600 m above sea level, the weight of the person is different. We need to calculate this difference and find the change in gravitational force.As we know, the gravitational force changes with altitude. The gravitational force acting on an object decreases as it moves farther away from the earth's center.To find the change in gravitational force, we need to first calculate the gravitational force acting on the person at sea level.
Gravitational force at sea level:F₁ = G × (Mm / R)²...[Equation 1]
Here, M is the mass of the earth, m is the mass of the person, R is the radius of the earth, and G is the gravitational constant. Putting the given values in Equation 1:F₁ = 6.674 × 10⁻¹¹ × (5.972 × 10²⁴ × 760) / (6.371 × 10⁶)²F₁ = 7.437 NNow, let's find the gravitational force acting on the person at 1600m above sea level.
Gravitational force at 1600m above sea level:F₂ = G × (Mm / (R+h))²...[Equation 2]Here, M is the mass of the earth, m is the mass of the person, R is the radius of the earth, h is the height of the airplane, and G is the gravitational constant. Putting the given values in Equation 2:F₂ = 6.674 × 10⁻¹¹ × (5.972 × 10²⁴ × 760) / (6.371 × 10⁶ + 1600)²F₂ = 7.333 NNow, we can find the change in gravitational force.ΔF = F₂ - F₁ΔF = 7.333 - 7.437ΔF = -0.104 NThe change in gravitational force is -0.104 N. A negative answer indicates a decrease in force.
Therefore, the gravitational force acting on the person has decreased by 0.104 N when they are on an airplane 1600 m above sea level as compared to the force measured at sea level.
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The binding energy for a particular metal is 0.576eV. What is the longest wavelength (in nm ) of light that can eject an electron from the metal's surface?.
When photons with energy more significant than the work function of a metal are exposed to a metal's surface, photoelectric emission occurs. longest wavelength of light that can eject an electron from the surface of a metal is 215 nm.
When light of a particular frequency is shone on a metal, the energy of each photon is transferred to the metal's electrons. As a result, electrons in the metal can overcome their bond's strength and leave the surface if they receive a sufficiently significant amount of energy. The wavelength (λ) of light that can eject electrons from a metal surface is determined by the metal's work function.
The maximum kinetic energy (Ek) of electrons emitted from a metal surface is determined by the difference between the energy of a photon (E) and the work function of a metal (Φ).The maximum kinetic energy of an electron is determined by the equation given below:Ek = E – Φwhere
E = Energy of the photonΦ = Work function of the metalTherefore, the longest wavelength of light that can eject an electron from the surface of a metal is determined by the following equation:λ = hc/EWhereh = Planck's constantc = Velocity of light E = Energy required to eject an electronλ = hc/ΦThe equation for the maximum kinetic energy of an electron isEk = hc/λ – Φ
Binding energy (Φ) for a particular metal = 0.576 eVThe velocity of light (c) = 3.00 x 10^8 m/sPlanck's constant (h) = 6.63 x [tex]10^{-34}[/tex]J/s We can use the formula below to convert electron-volts (eV) to joules (J).1 eV = 1.602 x [tex]10^{-19}[/tex] JΦ = 0.576 eV x 1.602 x [tex]10^{-19}[/tex] J/eVΦ = 9.22 x [tex]10^{-20}[/tex] Jλ = hc/Φ= (6.63 x [tex]10^{-34}[/tex] J/s) (3.00 x 10^8 m/s) / (9.22 x 10^-20 J)= 2.15 x [tex]10^{-7}[/tex] m= 215 nm
Therefore, the longest wavelength of light that can eject an electron from the surface of a metal is 215 nm.
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A circuit consists of a 256- resistor and a 0.191-H inductor. These two elements are connected in series across a generator that has a frequency of 115 Hz and a voltage of 351 V. (a) What is the current in the circuit? (b) Determine the phase angle between the current and the voltage of the generator. Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise.
a) The current in the circuit is 1.372 A.
b) The phase angle between the current and the voltage of the generator is 11.75°.
a) The current in the circuit is 1.372 A.
Step 1: The given values are: Resistance, R = 256 Ω Inductance, L = 0.191 HFrequency, f = 115 HzVoltage, V = 351 V
Step 2: Impedance of the circuit is given by the formula Z = √(R² + XL²),where XL = 2πfLZ = √(R² + (2πfL)²) = √(256² + (2π × 115 × 0.191)²) = 303.4 Ω
Step 3: The current in the circuit is given by the formula I = V/ZI = 351/303.4I = 1.372 A
b) The phase angle between the current and the voltage of the generator is 11.75°.Step 1: The phase angle between the current and the voltage of the generator is given by the formulaθ = tan⁻¹(XL/R)θ = tan⁻¹((2πfL)/R)θ = tan⁻¹((2π × 115 × 0.191)/256)θ = 11.75°.
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a) The current in the circuit is 1.372 A.
b) The phase angle between the current and the voltage of the generator is 11.75°.
a) The current in the circuit is 1.372 A.
Step 1: The given values are: Resistance, R = 256 Ω Inductance, L = 0.191 HFrequency, f = 115 HzVoltage, V = 351 V
Step 2: Impedance of the circuit is given by the formula Z = √(R² + XL²),where XL = 2πfLZ = √(R² + (2πfL)²) = √(256² + (2π × 115 × 0.191)²) = 303.4 Ω
Step 3: The current in the circuit is given by the formula I = V/ZI = 351/303.4I = 1.372 A
b) The phase angle between the current and the voltage of the generator is 11.75°.Step 1: The phase angle between the current and the voltage of the generator is given by the formulaθ = tan⁻¹(XL/R)θ = tan⁻¹((2πfL)/R)θ = tan⁻¹((2π × 115 × 0.191)/256)θ = 11.75°.
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A 37 kg box sits (is stationary) on an inclined plane that makes
an angle of 14° with the
horizontal. What is the minimum value of the coefficient of static
friction between the box
and the ramp?
The force of static friction must be equal to or greater than the component of weight along the incline. Therefore, Fs(max) >= mg * sin(θ)..
The weight of the box can be decomposed into two components: the force acting perpendicular to the plane (normal force) and the force acting parallel to the plane (component of weight along the incline). The normal force can be calculated as N = mg * cos(θ), where m is the mass of the box, g is the acceleration due to gravity, and θ is the angle of the inclined plane.
The force of static friction (Fs) acts parallel to the incline in the opposite direction to prevent the box from sliding. The maximum value of static friction can be given by Fs(max) = μs * N, where μs is the coefficient of static friction.
In order for the box to remain stationary, the force of static friction must be equal to or greater than the component of weight along the incline. Therefore, Fs(max) >= mg * sin(θ).
Substituting the values, we have μs * N >= mg * sin(θ).
By substituting N = mg * cos(θ), we have μs * mg * cos(θ) >= mg * sin(θ).
The mass (m) cancels out, resulting in μs * cos(θ) >= sin(θ).
Finally, we can solve for the minimum value of the coefficient of static friction by rearranging the inequality: μs >= tan(θ).
By substituting the given angle of 14°, the minimum value of the coefficient of static friction is μs >= tan(14°).
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A circular wire of radius 25 cm is oriented such that its plane is parallel to a 0.05 T magnetic field. The wire is rotated in 0.5 s such that its plane is perpendicular to the magnetic field. Determine the voltage generated in the wire.
The number of turns (N) in the wire loop is needed to calculate the voltage generated in the wire.
To determine the voltage generated in the wire, we can use Faraday's law of electromagnetic induction. According to the law, the induced voltage (emf) in a wire loop is given by the equation:
emf = -N * ΔΦ/Δt
Where:
- emf is the induced voltage (in volts, V).
- N is the number of turns in the wire loop.
- ΔΦ is the change in magnetic flux through the loop (in Weber, Wb).
- Δt is the time interval over which the change occurs (in seconds, s).
In this case, we are given:
- Radius of the circular wire = 25 cm = 0.25 m
- Magnetic field strength = 0.05 T
- Time interval = 0.5 s
- The wire is rotated from a position parallel to the magnetic field to a position perpendicular to it.
To find the change in magnetic flux (ΔΦ), we need to calculate the initial and final flux values and then find the difference between them.
Initial magnetic flux (Φi):
Φi = B * A_initial
Where B is the magnetic field strength and A_initial is the initial area of the wire loop.Since the wire loop is initially parallel to the magnetic field, the initial area (A_initial) is given by the formula for the area of a circle:
A_initial = π * (radius^2)
Final magnetic flux (Φf):
Φf = B * A_final
Where A_final is the final area of the wire loop when it is perpendicular to the magnetic field.The change in magnetic flux (ΔΦ) is then given by: ΔΦ = Φf - Φi
Finally, we can substitute the values into the formula for emf to find the voltage generated.
Let's calculate step by step:
1. Calculate the initial area (A_initial):
A_initial = π * (0.25 m)^2
2. Calculate the initial magnetic flux (Φi):
Φi = 0.05 T * A_initial
3. Calculate the final area (A_final):
A_final = π * (0.25 m)^2
4. Calculate the final magnetic flux (Φf):
Φf = 0.05 T * A_final
5. Calculate the change in magnetic flux (ΔΦ):
ΔΦ = Φf - Φi
6. Calculate the voltage (emf):
emf = -N * ΔΦ/Δt
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10. (10 points total) An object is placed 6 cm to the left of a converging lens. Its image forms 12 cm to the right of the lens. a) (3 points) What is focal length of the lens? b) (3 points) What is the magnification? c) (2 points) is the image upright, or inverted? (Please explain or show work.) d) (2 points) is the image real or virtilal? (Please explain or show work)
a) The focal length of the lens is 12 cm
b) The magnification is -2.
c) The magnification is negative (-2), meaning that the image is inverted.
d) Since the image distance is positive (12 cm to the right of the lens), it shows that the image is real.
How to determine the focal length of the lens?a) To evaluate the focal length of the lens, we shall use the lens formula:
1/f = 1/[tex]d_{0}[/tex] + 1/[tex]d_{i}[/tex]
where:
f = the focal length of the lens
d₀ = object distance
[tex]d_{i}[/tex] = image distance
Given:
d₀ = −6cm (since the object is 6 cm to the left of the lens),
[tex]d_{i}[/tex] = 12cm (the image forms is 12 cm to the right of the lens).
Putting the values:
1/f = 1/-6 + 1/12
We simplify:
1/f = 2/12 - 1/6
1/f = 1/12
Take the reciprocal of both sides:
f = 12cm
Therefore, the focal length of the lens is 12 cm.
b) The magnification (m) can be determined using the formula:
m = [tex]d_{i}[/tex] / [tex]d_{o}[/tex]
where:
[tex]d_{i}[/tex] = the object distance
[tex]d_{o}[/tex] = the image distance
Given:
[tex]d_{i}[/tex] = −6cm (object is 6 cm to the left of the lens),
[tex]d_{o}[/tex] = 2cm (since the image forms 12 cm to the right of the lens).
Plugging in the values:
m = -12/-6
m = -2
So, the magnification is -2.
c) The sign of the magnification tells us if the image is upright or inverted. In this situation, since the magnification is negative (-2), the image is inverted.
d) We shall put into account the sign of the image distance to determine if the image is real or virtual.
Here, the image distance is positive (12 cm to the right of the lens), indicating that the image is real.
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3. In a RL circuit, a switch closes at 0.0s. It has a battery with emf E=22V. At t=1.25s, the ammeter=0.2A. If R=40ohm, what is the magnetic energy when t=3.5s. Provide a response in J in the hundredth place. show all work.
The magnetic energy in the RL circuit when t=3.5s is 2.49 J. Which Provide a response in J in the hundredth place.
To find the magnetic energy in the RL circuit when t=3.5s, we need to calculate the current flowing through the circuit at that time and then use it to determine the energy stored in the inductor.
Given:
Emf of the battery (E) = 22V
Current at t=1.25s (I) = 0.2A
Resistance (R) = 40Ω
First, we need to find the inductance (L) of the circuit. Since the circuit contains only an inductor, the voltage across the inductor is equal to the emf of the battery. Therefore, we have:
E = L(dI/dt)
Rearranging the equation, we get:
L = E/(dI/dt)
The change in current with respect to time can be calculated as follows:
dI/dt = (I - I₀) / (t - t₀)
Where:
I₀ is the initial current at t₀ = 1.25s
Substituting the given values into the equation, we have:
dI/dt = (0.2A - I₀) / (3.5s - 1.25s)
Now, we can calculate the inductance (L):
L = 22V / [(0.2A - I₀) / (3.5s - 1.25s)]
Next, we need to calculate the energy stored in the inductor. The magnetic energy (W) is given by the equation:
W = (1/2) * L * I²
Substituting the known values, we have:
W = (1/2) * L * I²
Finally, substitute the values of L and I at t = 3.5s into the equation to find the magnetic energy at that time.
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Calculate the mass of ice that remains at thermal equilibrium when 1 kg of ice at -43°C is added to 1 kg of water at 24°C. Please report the mass of ice in kg to 3 decimal places. Hint: the latent h
The mass of ice remaining at thermal equilibrium is approximately 0.125 kg, assuming no heat loss or gain from the environment.
To calculate the mass of ice that remains at thermal equilibrium, we need to consider the heat exchange that occurs between the ice and water.
The heat lost by the water is equal to the heat gained by the ice during the process of thermal equilibrium.
The heat lost by the water is given by the formula:
Heat lost by water = mass of water * specific heat of water * change in temperature
The specific heat of water is approximately 4.186 kJ/(kg·°C).
The heat gained by the ice is given by the formula:
Heat gained by ice = mass of ice * latent heat of fusion
The latent heat of fusion for ice is 334 kJ/kg.
Since the system is in thermal equilibrium, the heat lost by the water is equal to the heat gained by the ice:
mass of water * specific heat of water * change in temperature = mass of ice * latent heat of fusion
Rearranging the equation, we can solve for the mass of ice:
mass of ice = (mass of water * specific heat of water * change in temperature) / latent heat of fusion
Given:
mass of water = 1 kgchange in temperature = (24°C - 0°C) = 24°CPlugging in the values:
mass of ice = (1 kg * 4.186 kJ/(kg·°C) * 24°C) / 334 kJ/kg
mass of ice ≈ 0.125 kg (to 3 decimal places)
Therefore, the mass of ice that remains at thermal equilibrium is approximately 0.125 kg.
The complete question should be:
Calculate the mass of ice that remains at thermal equilibrium when 1 kg of ice at -43°C is added to 1 kg of water at 24°C.
Please report the mass of ice in kg to 3 decimal places.
Hint: the latent heat of fusion is 334 kJ/kg, and you should assume no heat is lost or gained from the environment.
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A block of mass is attached to a spring with a spring constant and can move frictionlessly on a horizontal surface. The block is pulled out to the side a distance from the equilibrium position, and a starting speed is given to the left as it is released. Determine the maximum speed the block gets?
The maximum speed the block gets can be determined using the principle of conservation of mechanical energy. The maximum speed occurs when all potential energy is converted to kinetic energy.
When the block is pulled out to the side and released, it starts oscillating back and forth due to the restoring force provided by the spring. As it moves towards the equilibrium position, its potential energy decreases and is converted into kinetic energy. At the equilibrium position, all the potential energy is converted into kinetic energy, resulting in the maximum speed of the block.
According to the principle of conservation of mechanical energy, the total mechanical energy of the system (block-spring) remains constant throughout the motion. The mechanical energy is the sum of the potential energy (associated with the spring) and the kinetic energy of the block.
At the maximum speed, all the potential energy is converted into kinetic energy, so we can equate the potential energy at the starting position (maximum displacement) to the kinetic energy at the maximum speed. This gives us the equation:
(1/2)kx^2 = (1/2)mv^2
Where k is the spring constant, x is the maximum displacement from the equilibrium position, m is the mass of the block, and v is the maximum speed.
By rearranging the equation and solving for v, we can determine the maximum speed of the block.
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Caesar the Ape is in a tree, some height H above the ground. He sees his friend Will Rodman being chased by another angry Ape. Caesar grabs hold of a vine to swing down, grabs hold of Will Rodman, and swings up into another tree. Will the height Caesar swings up to with Rodman be higher than, the same as, or lower than the height Caesar starts from? Explain your reasoning using conservation laws of energy and/or momentum.
The height Caesar swings up to with Rodman will be lower than the height Caesar starts from.Conservation of energy and momentum play a significant role in determining the height to which Caesar swings up with Rodman. Energy and momentum are conserved when there is no external force acting on a system.
The law of conservation of energy states that the total energy in a closed system is constant, while the law of conservation of momentum states that the total momentum in a closed system is constant When he grabs hold of Will Rodman, he transfers some of his kinetic energy to him, causing the total kinetic energy of the system to remain constant.
The conservation of momentum states that the total momentum of the system is constant, which means that the combined momentum of Caesar and Will Rodman is the same before and after they swing.The total energy of the system is equal to the sum of the kinetic and potential energy. When Caesar and Will Rodman swing up into the second tree, some of their kinetic energy is converted back into potential energy, and their total energy is constant. As a result, the sum of their potential energy and kinetic energy at any point in the swing is the same as the sum of their potential energy and kinetic energy at any other point in the swing.
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A material has an index of refraction n = 1.47, the speed of the light in vacuum is c = 3 X 10^8. Which formula can be used to calculate the speed of the light in this material?
The speed of light in this material is approximately 2.04 x 10^8 meters per second. The formula that can be used to calculate the speed of light in a material is v = c/n.
The speed of light in vacuum is denoted by c, which has a constant value of approximately 3 x 10^8 meters per second. The index of refraction of a material is represented by n. To calculate the speed of light in the material, we divide the speed of light in vacuum (c) by the index of refraction (n).
Using the given values, we can substitute them into the formula:
v = c/n
= (3 x 10^8) / 1.47
≈ 2.04 x 10^8 meters per second
Therefore, the speed of light in this material is approximately 2.04 x 10^8 meters per second.
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Calculate the energy stored in a 750 F capacitor that has been charged to 12.0V.
The energy stored in the 750 F capacitor that has been charged to 12.0 V is 54,000 joules.
The energy stored in a capacitor can be calculated using the formula:
E = (1/2) * C * V^2
Where:
E is the energy stored in the capacitor
C is the capacitance of the capacitor
V is the voltage across the capacitor
Capacitance (C) = 750 F
Voltage (V) = 12.0 V
Substituting the values into the formula:
E = (1/2) * 750 F * (12.0 V)^2
Calculating the energy:
E = 0.5 * 750 F * 144 V^2
E = 54,000 J
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An astronaut whose mass is 105 kg has been working outside his spaceship, using a small, hand-held rocket gun to change his velocity in order to move around. After a while he finds that he has been careless: his gun is empty and he is out of reach of his
spaceship, drifting away from it at 0.7 m/s. The empty gun has a mass of 2.6 kg. How
can he get back to his ship? [A: throw it in the opposite direction with a v = 29 m/s]
To get back to his spaceship, the astronaut should throw the empty gun in the opposite direction with a velocity of 0.7 m/s.
To get back to his spaceship, the astronaut can use the principle of conservation of momentum. By throwing the empty gun in the opposite direction, he can change his momentum and create a force that propels him towards the spaceship.
Given:
Astronaut's mass (ma) = 105 kgAstronaut's velocity (va) = 0.7 m/sGun's mass (mg) = 2.6 kgGun's velocity (vg) = ?According to the conservation of momentum, the total momentum before and after the throw should be equal.
Initial momentum = Final momentum
(ma * va) + (mg * 0) = (ma * v'a) + (mg * v'g)
Since the gun is empty and has a velocity of 0 (vg = 0), the equation simplifies to:
ma * va = ma * v'a
The astronaut's mass and velocity remain the same before and after the throw, so we can solve for v'a.
va = v'a
Therefore, the astronaut needs to throw the empty gun with a velocity equal in magnitude but opposite in direction to his current velocity. So, he should throw the gun with a velocity of 0.7 m/s in the opposite direction (v'g = -0.7 m/s).
To calculate the magnitude of the velocity, we can use the equation:
ma * va = ma * v'a
105 kg * 0.7 m/s = 105 kg * v'a
v'a = 0.7 m/s
Therefore, the astronaut should throw the empty gun with a velocity of 0.7 m/s in the opposite direction (v'g = -0.7 m/s) to get back to his spaceship.
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Obtain the moment of inertia tensor of a thin uniform ring of
radius R, and mass M, with the origin of the coordinate system
placed at the center of the ring, and the ring lying in the
xy−plane.
The diagonal elements of the moment of inertia tensor are [tex]MR^2/2[/tex] for the x and y axes, and [tex]MR^2[/tex] for the z-axis. The moment of inertia tensor of a thin uniform ring can be obtained by considering its rotational symmetry and the distribution of mass.
The moment of inertia tensor (I) for a thin uniform ring of radius R and mass M, with the origin at the center of the ring and lying in the xy-plane, is given by I = [tex]M(R^2/2)[/tex] To derive the moment of inertia tensor, we need to consider the contributions of the mass elements that make up the ring. Each mass element dm can be treated as a point mass rotating about the z-axis.
The moment of inertia for a point mass rotating about the z-axis is given by I = [tex]m(r^2)[/tex], where m is the mass of the point and r is the perpendicular distance of the point mass from the axis of rotation.
In the case of a thin uniform ring, the mass is distributed evenly along the circumference of the ring. The perpendicular distance of each mass element from the z-axis is the same and equal to the radius R.
Since the ring has rotational symmetry about the z-axis, the moment of inertia tensor has off-diagonal elements equal to zero.
The diagonal elements of the moment of inertia tensor are obtained by summing the contributions of all the mass elements along the x, y, and z axes. Since the mass is uniformly distributed, each mass element contributes an equal amount to the moment of inertia along each axis.
Therefore, the diagonal elements of the moment of inertia tensor are [tex]MR^2/2[/tex] for the x and y axes, and [tex]MR^2[/tex] for the z-axis.
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Two convex thin lenses with focal lengths 12 cm and 18.0 cm aro aligned on a common avis, running left to right, the 12-сm lens being on the left. A distance of 360 сm separates the lenses. An object is located at a distance of 15.0 cm to the left of the 12-сm lens. A Make a sketch of the system of lenses as described above B. Where will the final image appear as measured from the 18-cm bens? Give answer in cm, and use appropriate sign conventions Is the final image real or virtual? D. is the famae upright or inverted? E What is the magnification of the final image?
The magnification is given by: M = v2/v1 = (54 cm)/(60 cm) = 0.9
This means that the image is smaller than the object, by a factor of 0.9.
A. Diagram B. Using the lens formula:
1/f = 1/v - 1/u
For the first lens, with u = -15 cm, f = +12 cm, and v1 is unknown.
Thus,1/12 = 1/v1 + 1/15v1 = 60 cm
For the second lens, with u = 360 cm - 60 cm = +300 cm, f = +18 cm, and v2 is unknown.
Thus,1/18 = 1/v2 - 1/300v2 = 54 cm
Thus, the image is formed at a distance of 54 cm to the right of the second lens, measured from its center, which makes it 54 - 18 = 36 cm to the right of the second lens measured from its right-hand side.
The image is real, as it appears on the opposite side of the lens from the object. It is inverted, since the object is located between the two lenses.
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A snow maker at a resort pumps 220 kg of lake water per minute and sprays it into the air above a ski run. The water droplets freeze in the air and fall to the ground, forming a layer of snow. If all of the water pumped into the air turns to snow, and the snow cools to the ambient air temperature of -6.8°C, how much heat does the snow-making process release each minute? Assume the temperature of the lake water is 13.9°C, and use 2.00x102)/(kg-Cº) for the specific heat capacity of snow
Find the amount of heat released each minute by using the following formula:Q = m × c × ΔT
where:Q = heat energy (in Joules or J),m = mass of the substance (in kg),c = specific heat capacity of the substance (in J/(kg·°C)),ΔT = change in temperature (in °C)
First, we need to find the mass of snow produced each minute. We know that 220 kg of water is pumped into the air each minute, and assuming all of it turns to snow, the mass of snow produced will be 220 kg.
Next, we can calculate the change in temperature of the water as it cools from 13.9°C to -6.8°C:ΔT = (-6.8°C) - (13.9°C)ΔT = -20.7°C
The specific heat capacity of snow is given as 2.00x102 J/(kg·°C), so we can substitute all the values into the formula to find the amount of heat released:Q = m × c × ΔTQ = (220 kg) × (2.00x102 J/(kg·°C)) × (-20.7°C)Q = -9.11 × 106 J
The snow-making process releases about 9.11 × 106 J of heat each minute.
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Undisturbed, the air has a pressure of 100 kPa (kiloPascals), otherwise known as atmospheric pressure. Sound waves are pressure waves where the pressure of the medium is slightly higher or lower than usual. They travel at 330 m/s in air. For the following, be sure to label your axes and intervals. A) Make a plot of pressure vs. time at one point in space for two complete cycles of a 400 Hz harmonic sound wave that has a maximum pressure 1 kPa above atomospheric pressure. B) Make a plot of pressure vs. distance at one point in time for this same wave over two cycles. C) On each of your plots in A) & B) use a dashed line (or a different color) to draw how the plot changes if the frequency is doubled to 800 Hz. D) On each of your plots in A) & B) use a dotted line (or a different color) to draw how the plot changes if the 400 Hz sound is made underwater, where the pressure is the same, but sound travels 5 time faster than in air.
A. The intervals on the x-axis would depend on the desired duration for two complete cycles of the sound wave.
B. The intervals on the x-axis would depend on the desired range of distance for two complete cycles of the sound wave.
C. On both plots A) and B), use a dashed line (or a different color) to represent the changes if the frequency is doubled to 800 Hz.
D. This would involve stretching the waves, resulting in a lower frequency and longer wavelength
A) Plot of pressure vs. time for two complete cycles of a 400 Hz harmonic sound wave:
The x-axis represents time, and the y-axis represents pressure. The pressure values on the y-axis would range from 99 kPa to 101 kPa to account for the maximum pressure 1 kPa above atmospheric pressure.
B) Plot of pressure vs. distance for the same wave over two cycles:
The x-axis represents distance, and the y-axis represents pressure. The pressure values on the y-axis would range from 99 kPa to 101 kPa to account for the maximum pressure 1 kP above atmospheric pressure.
C) This would involve compressing the waves, resulting in a higher frequency and shorter wavelength.
D) On both plots A) and B), use a dotted line (or a different color) to represent the changes if the 400 Hz sound is made underwater, where the pressure is the same but sound travels 5 times faster than in air. This would involve stretching the waves, resulting in a lower frequency and longer wavelength, while maintaining the same pressure levels.
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A free electron has a kinetic energy 19.4eV and is incident on a potential energy barrier of U = 34.5eV and width w=0.068mm. What is the probability for the electron to penetrate this barrier (in %)?
Tunneling of electrons, also known as quantum tunneling, is a phenomenon in quantum mechanics where particles, such as electrons, can penetrate through potential energy barriers that are classically forbidden.
The probability of penetration of an electron through a potential energy barrier can be determined using the expression: T= 16(E/U)(1 - E/U)exp (-2aw) where, a = sqrt (2m (U - E)) / h where T is the probability of penetration, E is the kinetic energy of the electron, U is the height of the potential energy barrier, w is the width of the barrier, m is the mass of the electron, and h is the Planck's constant.
The given values are, E = 19.4 eV, U = 34.5 eV, w = 0.068 mm = 6.8 × 10⁻⁵ cm, mass of the electron, m = 9.11 × 10⁻³¹ kg, and Planck's constant, h = 6.626 × 10⁻³⁴ J s. Substituting the given values, we get: a = sqrt (2 × 9.11 × 10⁻³¹ × (34.5 - 19.4) × 1.602 × 10⁻¹⁹) / 6.626 × 10⁻³⁴a = 8.26 × 10¹⁰ m⁻¹The probability of penetration: T= 16(19.4 / 34.5)(1 - 19.4 / 34.5)exp (-2 × 8.26 × 10¹⁰ × 6.8 × 10⁻⁵)T= 0.0255 or 2.55 %Therefore, the probability for the electron to penetrate this barrier is 2.55 %.
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A car travels at 50 km/hr for 2 hours. It then travels an additional distance of 23 km in 4 hour. The average speed of the car for the entire trip is(in km/hr),
Answer:
Distance travelled in 2 hours = 50 km/hr x 2 hrs = 100 km
Distance travelled in 4 hours = 23 km
Total distance travelled = 100 km + 23 km = 123 km
Total time taken = 2 hrs + 4 hrs = 6 hrs
Average speed = Total distance/Total time
Average speed = 123 km/6 hrs
Average speed = 20.5 km/hr
What is the time difference between an 30 year old on Earth and an 30 year old born and living her entire life in a spaceship and travelling at 0.89c ? Answer from each person's perspective.
The time difference From Person A's perspective on Earth, both individuals age 30 years. From Person B's perspective on the spaceship, only about 15 years pass.
According to special relativity, time dilation occurs when an object is moving at a significant fraction of the speed of light relative to another object. In this scenario, we have two individuals: a 30-year-old on Earth (let's call them Person A) and a 30-year-old who has spent their entire life on a spaceship traveling at 0.89 times the speed of light (let's call them Person B).
From Person A's perspective on Earth, time would pass normally for them. They would experience time in the same way as it occurs on Earth. So, if Person A remains on Earth, they would age 30 years.
From Person B's perspective on the spaceship, however, things would be different due to time dilation. When an object approaches the speed of light, time appears to slow down for that object relative to a stationary observer. In this case, Person B is traveling at 0.89 times the speed of light.
The time dilation factor, γ (gamma), can be calculated using the formula:
γ = 1 / √(1 - v^2/c^2)
where v is the velocity of the spaceship and c is the speed of light.
In this case, v = 0.89c. Plugging in the values, we get:
γ = 1 / √(1 - (0.89c)^2/c^2)
≈ 1.986
This means that, from Person B's perspective, time would appear to pass at approximately half the rate compared to Person A on Earth. So, if Person B spends 30 years on the spaceship, they would perceive only about 15 years passing.
It's important to note that this calculation assumes constant velocity and does not account for other factors like acceleration or deceleration. Additionally, the effects of time dilation become more significant as an object's velocity approaches the speed of light.
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TT 47. A transverse wave on a string is modeled with the wave function y(x, t) = (0.20 cm)sin (2.00 m-1x – 3.00 s-14+ 16). What is the height of the string with respect to the equilibrium position at a position x = 4.00 m and a time t = 10.00 s? =
The negative sign indicates that the height of the string at the given position and time is below the equilibrium position. Hence, the height is approximately -0.056 cm.
The height of the string with respect to the equilibrium position can be determined using the given wave function. At a position x = 4.00 m and a time t = 10.00 s, the wave function is y(x, t) = (0.20 cm)sin(2.00 m^(-1)x – 3.00 s^(-1)t + 16). By substituting the values of x and t into the wave function and evaluating the sine function, we can find the height of the string at that specific location and time.
The given wave function is y(x, t) = (0.20 cm)sin(2.00 m^(-1)x – 3.00 s^(-1)t + 16), where x represents the position along the string and t represents time. To find the height of the string at a specific position x = 4.00 m and time t = 10.00 s, we substitute these values into the wave function.
y(4.00 m, 10.00 s) = (0.20 cm)sin[2.00 m^(-1)(4.00 m) – 3.00 s^(-1)(10.00 s) + 16]
Simplifying the expression inside the sine function:
= (0.20 cm)sin[8.00 – 30.00 + 16]
= (0.20 cm)sin[-6.00]
Using the sine function, sin(-6.00) ≈ -0.279.
Therefore, y(4.00 m, 10.00 s) = (0.20 cm)(-0.279) ≈ -0.056 cm.
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You would like to use Ampere's law to find the Magnetic field a distance
r from a long straight wire. In order to take advantage of the symmetry
of the situation, the integration should be performed over:
In order to take advantage of the symmetry of the situation, the integration should be performed over a circular loop of radius r centered on the wire. The magnetic field will be tangential to the loop, and its magnitude will be proportional to the current in the wire and inversely proportional to the radius of the loop.
To take advantage of the symmetry of the situation and apply Ampere's law to find the magnetic field at a distance "r" from a long straight wire, you need to choose a closed path for integration that exhibits symmetry. In this case, the most suitable closed path is a circle centered on the wire and with a radius "r".
By choosing a circular path, the magnetic field will have a constant magnitude at every point on the path due to the symmetry of the wire. This allows us to simplify the integration and determine the magnetic field using Ampere's law.
The integration should be performed over the circular path with a radius "r" and centered on the wire.
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An object is 2m away from a convex mirror in a store, its image is 1 m behind the mirror. What is the focal length of the mirror? O 0.5 O -0.5 2 O-2
The focal length of the convex mirror in the store can be determined by using the mirror equation. The focal length of the mirror is -0.5m.
In the given scenario, the object is placed 2m away from a convex mirror, and the image is formed 1m behind the mirror. To find the focal length of the mirror, we can use the mirror equation:
1/f = 1/v - 1/u
where f is the focal length, v is the image distance, and u is the object distance.
Given that the image distance (v) is -1m (negative because it is formed behind the mirror) and the object distance (u) is -2m (negative because it is in front of the mirror), we can substitute these values into the mirror equation:
1/f = 1/-1 - 1/-2
Simplifying the equation gives:
1/f = -2/2 - 1/2
1/f = -3/2
f = -2/3
Therefore, the focal length of the convex mirror is approximately -0.5m.
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A parallel-plate capacitor has a charge Q and plates of area A . What force acts on one plate to attract it toward the other plate? Because the electric field between the plates is E=Q / A €₀, you might think the force is F=Q E=Q²/ A €₀ This conclusion is wrong because the field E includes contributions from both plates, and the field created by the positive plate cannot exert any force on the positive plate. Show that the force exerted on each plate is actually F= Q² / 2 A€₀ . Suggestion: Let C = €₀A / x for an arbitrary plate separation x and note that the work done in separating the two charged plates is W = in F d x .
To show that the force exerted on each plate of a parallel-plate capacitor is F=Q²/2A€₀, we can follow the suggested approach.
Let's start with the equation W = F * dx, where W is the work done, F is the force, and dx is the separation between the plates. The work done in separating the two charged plates can be expressed as W = (1/2)C(V^2), where C is the capacitance and V is the potential difference between the plates. Since C = €₀A / x, we can substitute it into the equation to get W = (1/2)(€₀A / x)(V^2).
The potential difference V can be written as V = Q / (€₀A), where Q is the charge on one of the plates.
Substituting V into the equation, we have W = (1/2)(€₀A / x)((Q / (€₀A))^2).
Simplifying the equation further, W = (1/2)(Q^2 / (€₀A)(x)).
Since W = F * dx, we can equate the two equations to get (1/2)(Q^2 / (€₀A)(x)) = F * dx.
Dividing both sides by dx and rearranging, we obtain F = (1/2)(Q^2 / (€₀A)(x)).
Now, since A and €₀ are constant for a given capacitor, we can simplify the equation to F = Q^2 / (2A€₀x).
Therefore, the force exerted on each plate of a parallel-plate capacitor is F = Q^2 / (2A€₀), as required.
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A mass is placed on a frictionless, horizontal table. A spring (k=110 N/m), which can be stretched or compressed, is placed on the table. A 3-kg mass is anchored to the wall. The equilibrium position is marked at zero. A student moves the mass out to x=7.0 cm and releases it from rest. The mass oscillates in simple harmonic motion. Find the position, velocity, and acceleration of the mass at time t=3.00 s.
Position of the mass after t=3.00 s = 0.0638 m ; Velocity of the mass after t=3.00 s= -0.436 m/s ; Acceleration of the mass after t=3.00 s = -2.98 m/s².
Step 1: Calculate the angular frequencyω = √(k/m), where k is the spring constant and m is the mass.ω = √(110/3)
= 6.83 rad/s
Step 2: Determine the amplitude of oscillation
the displacement equation x(t) = A cos(ωt + φ), where A is the amplitude of oscillation, and φ is the phase constant. x(0) = A cos(φ)
At equilibrium position, x(0) = 0, so A cos(φ) = 0, implying that A = 0 as cos(φ) cannot be zero.
Therefore, the mass does not oscillate at the equilibrium position.
Step 3: Calculate the phase constant φ = cos⁻¹(x(0) / A)
At time t = 0, the mass is at x = 7.0 cm,
sox(0) = 7.0 cm
= 0.07 m
Using x(t) = A cos(ωt + φ),0.07 m
= A cos(φ)cos(φ)
= 0.07/Aφ
= cos⁻¹(0.07/A)
For simplicity, assume that the mass is released from x = 7.0 cm at t = 0 and moves towards the equilibrium position x = 0. Since the phase constant is zero at the equilibrium position, the value of the phase constant is 0 for all subsequent instants.
Step 4: Calculate the position of the mass x(t) = A cos(ωt)
The position of the mass at t = 3.00 s is,
x(3.00 s) = A cos(ωt)
= 0.0638 m.
Step 5: Calculate the velocity of the mass v(t) = -Aω sin(ωt)
The velocity of the mass at t = 3.00 s is,
v(3.00 s) = -0.436 m/s.
Step 6: Calculate the acceleration of the mass
a(t) = -Aω2 cos(ωt)
The acceleration of the mass at t = 3.00 s is,
a(3.00 s) = -2.98 m/s²
Position of the mass after t=3.00 s: x(3.00 s)
= 0.0638 m
Velocity of the mass after t=3.00 s: v(3.00 s)
= -0.436 m/s
Acceleration of the mass after t=3.00 s: a(3.00 s)
= -2.98 m/s².
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17. A 1.5 kg object moves as a function of time as X(t) = 5 cos [3t+2.5), where x is in meter and t in second. What is the kinetic energy of the object at t=4s? (5) (a) 174 ) (b) 147) (c) 417) (d) 741 18. An aluminum rod is heated from 20°C to 100°C. The final length of the rod is 50 cm. what is the change in length of the rod? [The coefficient of linear expansion of the rod is 24 x 10^/C] (5) (a) 0.01 cm (b) 0.1 cm (c) 0.1 mm (d) 0.02 cm 19. What is the amount of heat required to change 50 g of ice at -20°C to water at 50°C? [Specific heat capacity of ice =0.5 calg, Specfic heat capacity of water = 1 cal/gºC. Latent heat of fusion of ice = 79.6 cal/g] (5) (a) 6089 cal (b) 6980 cal (c) 6890 cal (d) 6098 cal 20. What is the r.m.s. speed of the Nitrogen molecule at 50C? [M = 28 g/mol; NA=6.023 x 10 molecules/mol] (5) (a) 534.6 m/s (b) 536.4 m's (c) 364.5 m/s (d) 465.3 m/s
The kinetic energy of the object at t = 4s is approximately 133.87 J. The change in length of the rod is 0.096 cm. The amount of heat required to change 50 g of ice at -20°C to water at 50°C is 7480 cal. The rms speed of the Nitrogen molecule at 50°C is approximately 465.3 m/s.
17. To find the kinetic energy of the object at t = 4s, we can differentiate the given position function with respect to time to obtain the velocity function and then calculate the kinetic energy using the formula KE = (1/2)mv^2.
Given: X(t) = 5cos(3t + 2.5), where x is in meters and t is in seconds.
Differentiating X(t) with respect to t:
V(t) = -15sin(3t + 2.5)
At t = 4s:
V(4) = -15sin(3(4) + 2.5)
V(4) ≈ -13.73 m/s (rounded to two decimal places)
Now, we can calculate the kinetic energy:
KE = (1/2)(1.5 kg)(-13.73 m/s)^2
KE ≈ 133.87 J (rounded to two decimal places)
Therefore, the kinetic energy of the object at t = 4s is approximately 133.87 J.
18. The change in length (ΔL) of the rod can be calculated using the formula ΔL = αLΔT, where α is the coefficient of linear expansion, L is the initial length of the rod, and ΔT is the change in temperature.
Given: Coefficient of linear expansion (α) = 24 x 10^-6 /°C, Initial length (L) = 50 cm, Change in temperature (ΔT) = (100°C - 20°C) = 80°C.
ΔL = (24 x 10^-6 /°C)(50 cm)(80°C)
ΔL = 0.096 cm
Therefore, the change in length of the rod is 0.096 cm.
19. To calculate the amount of heat required, we need to consider the phase changes and temperature changes separately.
First, we need to heat the ice from -20°C to its melting point:
Heat = mass × specific heat capacity × temperature change
Heat = 50 g × 0.5 cal/g°C × (0°C - (-20°C))
Heat = 1000 cal
Next, we need to melt the ice at 0°C:
Heat = mass × latent heat of fusion
Heat = 50 g × 79.6 cal/g
Heat = 3980 cal
Finally, we need to heat the water from 0°C to 50°C:
Heat = mass × specific heat capacity × temperature change
Heat = 50 g × 1 cal/g°C × (50°C - 0°C)
Heat = 2500 cal
Total heat required = 1000 cal + 3980 cal + 2500 cal = 7480 cal
Therefore, the amount of heat required to change 50 g of ice at -20°C to water at 50°C is 7480 cal.
20. The root mean square (rms) speed of a molecule can be calculated using the formula vrms = √(3kT/m), where k is the Boltzmann constant, T is the temperature in Kelvin, and m is the molar mass of the gas.
Given: Temperature (T) = 50°C = 323 K, Molar mass (M) = 28 g/mol.
First, convert the molar mass from grams to kilograms:
M = 28 g/mol = 0.028 kg/mol
Now, we can calculate the rms speed:
vrms = √(3kT/m)
vrms = √[(3 × 1.38 × 10^-23 J/K) × 323 K / (0.028 kg/mol)]
vrms ≈ 465.3 m/s (rounded to one decimal place)
Therefore, the rms speed of the Nitrogen molecule at 50°C is approximately 465.3 m/s.
Learn more about kinetic energy: https://brainly.com/question/30337295
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