The contestant calculates the resultant displacement by adding the three given displacements vectorially.
To determine the location of the buried keys, the contestant needs to calculate the resultant displacement by adding the three given displacements together. Here's how she can calculate it:
1. Start by converting the given displacements into their respective vector form. Each vector can be represented as a combination of horizontal (x) and vertical (y) components.
For the first displacement:
Magnitude: 72.4 m
Direction: 32.0° east of north
To find the horizontal and vertical components, we can use trigonometric functions. The eastward component can be found using cosine, and the northward component can be found using sine.
Horizontal component: 72.4 m * cos(32.0°)
Vertical component: 72.4 m * sin(32.0°)
For the second displacement:
Magnitude: 57.3 m
Direction: 36.0° south of west
To find the horizontal and vertical components, we use the same approach:
Horizontal component: 57.3 m * cos(180° - 36.0°) [180° - 36.0° is used because it's south of west]
Vertical component: 57.3 m * sin(180° - 36.0°)
For the third displacement:
Magnitude: 17.8 m
Direction: Straight south
The horizontal component for this displacement is 0 since it's purely vertical, and the vertical component is simply -17.8 m (negative because it's south).
2. Add up the horizontal and vertical components separately for all three displacements:
Total horizontal component = Horizontal component of displacement 1 + Horizontal component of displacement 2 + Horizontal component of displacement 3
Total vertical component = Vertical component of displacement 1 + Vertical component of displacement 2 + Vertical component of displacement 3
3. Calculate the magnitude and direction of the resultant displacement using the total horizontal and vertical components:
Resultant magnitude = √(Total horizontal component^2 + Total vertical component^2)
Resultant direction = arctan(Total vertical component / Total horizontal component)
The contestant needs to calculate these values to determine the location where the keys to the new Porsche are buried.
The complete question should be:
The three finalists in a contest are brought to the center of a large, flat field. Each is given a meter stick, a compass, a calculator, a shovel, and (in a different order for each contestant) the following three displacements:
72.4 m, 32.0° east of north; 57.3 m, 36.0° south of west;17.8 m straight south.The three displacements lead to the point where the keys to a new Porsche are buried. Two contestants start measuring immediately, but the winner first calculates where to go. What does she calculate?
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A closed box is filled with dry ice at a temperature of -91.7 °C, while the outside temperature is 29.2 °C. The box is cubical, measuring 0.284 m on a side, and the thickness of the walls is 3.62 x 102 m. In one day, 3.02 x 106 J of heat is conducted through the six walls. Find the thermal conductivity of the material from which the box is made.
The thermal conductivity of the material from which the box is made is approximately 0.84 W/(m·K).
The heat conducted through the walls of the box can be determined using the formula:
Q = k * A * (ΔT / d)
Where:
Q is the heat conducted through the walls,
k is the thermal conductivity of the material,
A is the surface area of the walls,
ΔT is the temperature difference between the inside and outside of the box, and
d is the thickness of the walls.
Given that the temperature difference ΔT is (29.2 °C - (-91.7 °C)) = 121.7 °C and the heat conducted Q is 3.02 x [tex]10^{6}[/tex] J, we can rearrange the formula to solve for k:
k = (Q * d) / (A * ΔT)
The surface area A of the walls can be calculated as:
A = 6 * [tex](side length)^{2}[/tex]
Substituting the given values, we have:
A = 6 * (0.284 m)2 = 0.484 [tex]m^{2}[/tex]
Now we can substitute the values into the formula:
k = (3.02 x [tex]10^{6}[/tex] J * 3.62 x [tex]10^{-2}[/tex] m) / (0.484 [tex]m^{2}[/tex] * 121.7 °C)
Simplifying the expression, we find:
k = 0.84 W/(m·K)
Therefore, the thermal conductivity of the material from which the box is made is approximately 0.84 W/(m·K).
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Special Relativity 11. Two meteors of rest-mass 0.1 and 0.2 kg, respectively, collide. If the relative speed before collision is 0.1 c and an observer sees them coming with equal and opposite speed and sees the lighter meteor go off at right angles to the original direction of motion, what will be the deflection of the heavier meteor according to the observer? How will this process appear to an observer comoving with: (i) the heavier meteor; and (ii) the lighter meteor? How will it appear to the centre of mass observer? Please solve all parts i.e, 1)Deflection of heavier meteor according to the observer 2) How this process looks to an observer Comoving with the heavier meteor 3) How this process looks to an observer Comoving with the lighter meteor 4)How will it appear to the centre of mass observer Special Relativity 11. Two meteors of rest-mass 0.1 and 0.2 kg, respectively, collide. If the relative speed before collision is 0.1 c and an observer sees them coming with equal and opposite speed and sees the lighter meteor go off at right angles to the original direction of motion, what will be the deflection of the heavier meteor according to the observer? How will this process appear to an observer comoving with: (i) the heavier meteor; and (ii) the lighter meteor? How will it appear to the centre of mass observer? Please solve all parts i.e, 1)Deflection of heavier meteor according to the observer 2) How this process looks to an observer Comoving with the heavier meteor 3) How this process looks to an observer Comoving with the lighter meteor 4)How will it appear to the centre of mass observer
The Lorentz factor for a speed of 0.1 c is 1.005, so the deflection of the heavier meteor is 1.005. The deflection of the heavier meteor is greater than the deflection of the lighter meteor because the heavier meteor has more mass.
1. Deflection of heavier meteor according to the observer
The deflection of the heavier meteor is given by the following equation:
deflection = (gamma - 1) * sin(theta)
where:
gamma is the Lorentz factor, given by:
gamma = 1 / sqrt(1 - v^2 / c^2)
v is the speed of the meteor, given by:
v = 0.1 c
theta is the angle between the direction of motion of the meteor and the direction of the deflection.
In this case, theta is 90 degrees, so the deflection is:
deflection = (gamma - 1) * sin(90 degrees) = gamma
The Lorentz factor for a speed of 0.1 c is 1.005, so the deflection of the heavier meteor is 1.005.
2. How this process looks to an observer comoving with the heavier meteor
To an observer comoving with the heavier meteor, the lighter meteor would appear to come from the side and collide with the heavier meteor head-on. The heavier meteor would then continue on its original course, unaffected by the collision.
3. How this process looks to an observer comoving with the lighter meteor
To an observer comoving with the lighter meteor, the heavier meteor would appear to come from the front and collide with the lighter meteor from behind. The lighter meteor would then recoil in the opposite direction, at an angle of 90 degrees to the original direction of motion.
4. How will it appear to the center of mass observer
To the center of the mass observer, the two meteors would appear to collide head-on. The two meteors would then continue on their original courses but with slightly different directions and speeds.
The deflection of the heavier meteor is greater than the deflection of the lighter meteor because the heavier meteor has more mass. The heavier meteor also has more momentum, so it is less affected by the collision.
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If a planetary nebula expands at a rate of 35 km s and has a lifetime of 25900 years, how large will it grow before it stops being visible? maximum radius of planetary nebula _____ km
The maximum radius of the planetary nebula is approximately 28.5 billion kilometers (rounded to two significant figures).
The maximum radius of a planetary nebula can be determined using the relation:radius = speed x age of nebula
In this case, the planetary nebula expands at a rate of 35 km/s and has a lifetime of 25900 years.
Therefore, the maximum radius of planetary nebula is calculated as follows:
radius = speed x age of nebula= 35 km/s x 25900 years (Note that the units of years need to be converted to seconds)
1 year = 365 days = 24 hours/day = 60 minutes/hour = 60 seconds/minute
Thus, 25900 years = 25900 x 365 x 24 x 60 x 60 seconds= 816336000 seconds
Plugging in the values, we get:
radius = 35 km/s x 816336000 s= 28521760000 km
Therefore, the maximum radius of the planetary nebula is approximately 28.5 billion kilometers (rounded to two significant figures).
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ASK YOUR TEACHER PRACTICE ANOTH The velocity of a proton in an accelerator is known to an accuracy of 0.211% of the speed of light (This could be small compared with its velocity) What is the smallest possible uncertainty in its position in m)? Additional Material
The correct answer is the smallest possible uncertainty in the position of the proton is 5.73 × 10-14 m.
According to the Heisenberg uncertainty principle, it is impossible to simultaneously know the precise position and momentum of an object at the same time. Thus, a finite uncertainty will always exist in both quantities. As a result, the minimum uncertainty in the position of the proton can be estimated using the following formula: Δx × Δp ≥ h/2π where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant (6.626 × 10-34 J · s).
The uncertainty in momentum can be calculated as follows:Δp = mv × Δv where m is the mass of the proton, v is its velocity, and Δv is the uncertainty in velocity.Δv = 0.211% of the speed of light = 2.17 × 105 m/s (Given)
Thus, Δp = mv × Δv= 1.67 × 10-27 kg × 2.17 × 105 m/s= 3.63 × 10-22 kg · m/s
Therefore,Δx × Δp = h/2πΔx = (h/2π) / Δp= (6.626 × 10-34 J · s / 2π) / 3.63 × 10-22 kg · m/s= 5.73 × 10-14 m
Thus, the smallest possible uncertainty in the position of the proton is 5.73 × 10-14 m.
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Review. A string is wound around a uniform disk of radius R and mass M . The disk is released from rest with the string vertical and its top end tied to a fixed bar (Fig. P10.73). Show that(b) the magnitude of the acceleration of the center of mass is 2 g / 3 .
Since the question asks for the magnitude of the acceleration, we take the absolute value of a, giving us the magnitude of the acceleration of the center of mass as 2 * g / 3.
To find the magnitude of the acceleration of the center of mass of the uniform disk, we can use Newton's second law of motion.
1. Let's start by considering the forces acting on the disk. Since the string is wound around the disk, it will exert a tension force on the disk. We can also consider the weight of the disk acting vertically downward.
2. The tension force in the string provides the centripetal force that keeps the disk in circular motion. This tension force can be calculated using the equation T = m * a,
3. The weight of the disk can be calculated using the equation W = m * g, where W is the weight, m is the mass of the disk, and g is the acceleration due to gravity.
4. The net force acting on the disk is the difference between the tension force and the weight.
5. Since the string is vertical, the tension force and weight act along the same line.
6. Substituting the equations, we have m * a - m * g = m * a.
7. Simplifying the equation, we get -m * g = 0.
8. Solving for a, we find a = -g.
9. Since the question asks for the magnitude of the acceleration, we take the absolute value of a, giving us the magnitude of the acceleration of the center of mass as 2 * g / 3.
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What is the diameter of a 1.50 m length of tungsten wire whose resistance is 0.440 12? Submit Answer Tries 0/10
To determine the diameter of a tungsten wire, we can use the formula for resistance:
R = (ρ * L) / A
where R is the resistance, ρ is the resistivity of tungsten, L is the length of the wire, and A is the cross-sectional area of the wire.
The resistivity of tungsten (ρ) is approximately 5.6 x 10^-8 ohm-meters.
Let's rearrange the formula to solve for the cross-sectional area (A):
A = (ρ * L) / R
A = (5.6 x 10^-8 ohm-meters * 1.50 meters) / 0.44012 ohms
A = 1.9081 x 10^-7 square meters
The area of a circle is given by the formula:
A = π * (d/2)^2
where d is the diameter of the wire.
Let's rearrange this formula to solve for the diameter (d):
d = √((4 * A) / π)
d = √((4 * 1.9081 x 10^-7 square meters) / π)
d ≈ 2.779 x 10^-4 meters
To convert the diameter from meters to millimeters (mm), multiply by 1000:
d ≈ 2.779 x 10^-1 mm
So, the diameter of the tungsten wire is approximately 0.2779 mm.
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1. The heaviest bench press a person can complete is 200 lbs. What percentage of their maximum are they lifting if they exercise with 140 lbs?
2. A person is lowering a barbell during a bench press exercisE. If upward motion is defined as positive, what can be said about the vertical velocity of the bar?
a. zero
b. not enough information to answer
c. it is positive
d. it is negative
3. Speeds in meters per second can be converted to miles per hour since one m/s equals 2.24 mph. How fast in mph is a volleyball spike with a speed of 30 m/s?
A person lifting 140 lbs in a bench press is lifting 70% of their maximum weight.
To determine the percentage of their maximum weight, we divide the weight being lifted (140 lbs) by the maximum weight (200 lbs) and multiply by 100. Therefore, (140/200) * 100 = 70%. So, when exercising with 140 lbs, the person is lifting 70% of their maximum weight.
Regarding the vertical velocity of the barbell during a bench press exercise, since the person is lowering the barbell, the motion is in the downward direction.
If upward motion is defined as positive, the vertical velocity of the barbell would be negative. The negative sign indicates the downward direction, indicating that the barbell is moving downward during the exercise.
To convert the speed of a volleyball spike from meters per second (m/s) to miles per hour (mph), we can use the conversion factor of 1 m/s = 2.24 mph.
Given that the spike speed is 30 m/s, we can multiply this value by the conversion factor: 30 m/s * 2.24 mph = 67.2 mph. Therefore, the volleyball spike has a speed of 67.2 mph.
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Calculate the moment of inertia of a plate of side 10 cm (square)
and mass 0.2 kg.
The moment of inertia of a plate with side length 10 cm and mass 0.2 kg is 0.0083 kg·m².
The moment of inertia of a rectangular plate about an axis passing through its center and perpendicular to its plane can be calculated using the formula: I = (1/12) * m * (a² + b²), where I is the moment of inertia, m is the mass of the plate, and a and b are the side lengths of the plate.
In this case, since the plate is a square, both side lengths are equal to 10 cm. Substituting the values into the formula, we have I = (1/12) * 0.2 kg * (0.1 m)² = 0.0083 kg·m².
Therefore, the moment of inertia of the given plate is 0.0083 kg·m².
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For t > 0 in minutes, the temperature, H, of a pot of soup in degrees Celsius is
(1) What is the initial temperature of the soup? (2) Find the value of # '(10) with UNITS. Explain its meaning in terms of
the temperature of the soup.
Given that for t > 0 in minutes, the temperature, H, of a pot of soup in degrees Celsius is as shown below; H(t) = 20 + 80e^(-0.05t). (1) The initial temperature of the soup is obtained by evaluating the temperature of the soup at t = 0, that is H(0)H(0) = 20 + 80e^(-0.05(0))= 20 + 80e^0= 20 + 80(1)= 20 + 80= 100°C. The initial temperature of the soup is 100°C.
(2) The derivative of H(t) with respect to t is given by H'(t) = -4e^(-0.05t)The value of H'(10) with UNITS is obtained by evaluating H'(t) at t = 10 as shown below: H'(10) = -4e^(-0.05(10))= -4e^(-0.5)≈ -1.642°C/minute. The value of H'(10) with UNITS is -1.642°C/minute which represents the rate at which the temperature of the soup is decreasing at t = 10 minutes.
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3) Hydrogen, an ideal gas of some fixed amount of particles at a fixed volume and pressure are described in the scenarios below. The mass of a hydrogen atom is 1.67 10-27 kg, and the Boltzmann constant is 1.38 x 10-23 J/K. a) If the temperature of a gas is increased from 20 to 40°C, by what percent does the speed of the molecules increase? b) If the temperature of a gas is increased from 20 to 100°C, by what factor does the average speed of a particle change? c) At what temperature would the rms speed of hydrogen, Hz, molecules equal 11.2 km/s?
(a)The speed of the molecules increases by 100%. (b) The average speed of a particle changes by a factor of 5 . (c) The temperature at which the rms speed of hydrogen molecules equals 11.2 km/s is approximately 8.063 K.
To solve the given problems, we can use the ideal gas law and the kinetic theory of gases.
(a) To calculate the percent increase in the speed of molecules when the temperature is increased from 20 to 40°C, we can use the formula for the average kinetic energy of gas molecules:
Average kinetic energy = (3/2) * k * T
The average kinetic energy is directly proportional to the temperature. Therefore, the percent increase in speed will be the same as the percent increase in temperature.
Percent increase = ((new temperature - old temperature) / old temperature) * 100%
Percent increase = ((40°C - 20°C) / 20°C) * 100%
Percent increase = 100%
Therefore, the speed of the molecules increases by 100%.
(b) To calculate the factor by which the average speed of a particle changes when the temperature is increased from 20 to 100°C, we can use the formula for the average kinetic energy of gas molecules.
Average kinetic energy = (3/2) * k * T
The average kinetic energy is directly proportional to the temperature. Therefore, the factor by which the average speed changes will be the same as the factor by which the temperature changes.
Factor change = (new temperature / old temperature)
Factor change = (100°C / 20°C)
Factor change = 5
Therefore, the average speed of a particle changes by a factor of 5.
(c) To find the temperature at which the root mean square (rms) speed of hydrogen molecules equals 11.2 km/s, we can use the formula for rms speed:
rms speed = sqrt((3 * k * T) / m)
Rearranging the formula:
T = (rms speed)^2 * m / (3 * k)
Plugging in the given values:
T = (11.2 km/s)^2 * (1.67 x 10^-27 kg) / (3 * 1.38 x 10^-23 J/K)
T = (11.2 * 10^3 m/s)^2 * (1.67 x 10^-27 kg) / (3 * 1.38 x 10^-23 J/K)
T = (1.2544 x 10^5 m²/s²) * (1.67 x 10^-27 kg) / (4.14 x 10^-23 J/K)
T ≈ 8.063 K
Therefore, the temperature at which the rms speed of hydrogen molecules equals 11.2 km/s is approximately 8.063 K.
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4. Measurements indicate that an atom remains in an excited state for an average time of 50.0 ns before making a transition to the ground state with the simultaneous emission of a 2.1-eV photon. (a) Estimate the uncertainty in the frequency of the photon. (b) What fraction of the photon's average frequency is this? 5. Suppose an electron is confined to a region of length 0.1 nm (of the order of the size of a hydrogen atom). (a) What is the minimum uncertainty of its momentum? (b) What would the uncertainty in momentum be if the confined length region doubled to 0.2 nm ?
4. The uncertainty in the frequency of a photon is estimated using the energy-time uncertainty principle, fraction of the photon's average frequency cannot be determined.
5. The minimum uncertainty in momentum is calculated using the position-momentum uncertainty principle, and when the confined length region doubles, the uncertainty in momentum also doubles.
4. (a) To estimate the uncertainty in the frequency of the photon, we can use the energy-time uncertainty principle:
ΔE Δt ≥ ħ/2
where ΔE is the uncertainty in energy, Δt is the uncertainty in time, and ħ is the reduced Planck's constant.
The uncertainty in energy is given by the energy of the photon, which is 2.1 eV. We need to convert it to joules:
1 eV = 1.6 × 10^−19 J
2.1 eV = 2.1 × 1.6 × 10^−19 J
ΔE = 3.36 × 10^−19 J
The average time is 50.0 ns, which is 50.0 × 10^−9 s.
Plugging the values into the uncertainty principle equation, we have:
ΔE Δt ≥ ħ/2
(3.36 × 10^−19 J) Δt ≥ (ħ/2)
Δt ≥ (ħ/2) / (3.36 × 10^−19 J)
Δt ≥ 2.65 × 10^−11 s
Now, to find the uncertainty in frequency, we use the relationship:
ΔE = Δhf
where Δh is the uncertainty in frequency.
Δh = ΔE / f
Substituting the values:
Δh = (3.36 × 10^−19 J) / f
To estimate the uncertainty in frequency, we need to know the value of f.
(b) To find the fraction of the photon's average frequency, we divide the uncertainty in frequency by the average frequency:
Fraction = Δh / f_average
Since we don't have the value of f_average, we can't calculate the fraction without additional information.
5. (a) The minimum uncertainty in momentum (Δp) can be calculated using the position-momentum uncertainty principle:
Δx Δp ≥ ħ/2
where Δx is the uncertainty in position.
The confined region has a length of 0.1 nm, which is 0.1 × 10^−9 m.
Plugging the values into the uncertainty principle equation, we have:
(0.1 × 10^−9 m) Δp ≥ ħ/2
Δp ≥ (ħ/2) / (0.1 × 10^−9 m)
Δp ≥ 5 ħ × 10^9 kg·m/s
(b) If the confined length region doubles to 0.2 nm, the uncertainty in position doubles as well:
Δx = 2(0.1 × 10^−9 m) = 0.2 × 10^−9 m
Plugging the new value into the uncertainty principle equation, we have:
(0.2 × 10^−9 m) Δp ≥ ħ/2
Δp ≥ (ħ/2) / (0.2 × 10^−9 m)
Δp ≥ 2.5 ħ × 10^9 kg·m/s
Therefore, the uncertainty in momentum doubles when the confined length region doubles.
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A defective starter motor draws 285 AA from a car’s 12.6-VV battery, dropping the voltage at the battery terminals to 7.33 VV. A good starter motor should draw only 112 AA.
Find the battery terminal voltage with a good starter
A good starter motor drawing a current of 112 A, the battery's terminal voltage would be around 4.944 V.
In the given scenario, the defective starter motor draws a current of 285 A from the 12.6 V battery, resulting in a voltage drop at the battery terminals to 7.33 V. On the other hand, a good starter motor should draw only 112 A.
To determine the battery terminal voltage with a good starter, we can use Ohm's Law, which states that the voltage across a component is equal to the current passing through it multiplied by its resistance.
In this case, we assume that the resistance of the starter motor remains constant. We can set up a proportion using the current values for the defective and good starter motors:
V = I R
285 A / 12.6 V = 112 A / x V
285 A * x V = 12.6 V * 112 A
x V = (12.6 V * 112 A) / 285 A
x V ≈ 4.944 V
Therefore, the battery terminal voltage with a good starter motor would be approximately 4.944 V.
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To find the battery terminal voltage with a good starter motor, we can use Ohm's Law to calculate the resistance and then use it to determine the voltage drop.
Explanation:To find the battery terminal voltage with a good starter, we can use Ohm's Law, which states that voltage (V) is equal to current (I) multiplied by resistance (R). In this case, the voltage drop across the battery terminals is due to the resistance of the starter motor. We can calculate the resistance using the formula R = V/I. For the defective starter motor, the resistance would be 12.6 V / 285 A = 0.0442 ohm. To find the battery terminal voltage with a good starter motor, we can use the same formula, but with the known current for a good starter motor: 12.6 V / 112 A = 0.1125 ohm. Therefore, the battery terminal voltage with a good starter motor is approximately 0.1125 V.
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C12: You have a circuit made of two 1.5 V batteries connected together (that's two AA batteries) and a 3.3k resistor. What is the current through the resistor and the voltage drop across the resistor?
The current through the resistor in the circuit is approximately 0.909 mA, and the voltage drop across the resistor is approximately 3.00 V.
In the given circuit, we have two 1.5 V batteries connected in series, resulting in a total voltage of 3 V. The resistor has a value of 3.3 kΩ.
To calculate the current through the resistor, we can use Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R). Therefore,
[tex]I=\frac{V}{R}[/tex]
Substituting the values, we get [tex]I=\frac{3V}{3.3 k\Omega}=0.909 mA[/tex].
Since the batteries are connected in series, the current passing through the resistor is the same as the total circuit current.
To find the voltage drop across the resistor, we can use Ohm's Law again [tex]V=IR[/tex].
Substituting the values, we get [tex]V=0.909mA \times 3.3k\Omega=3.00V.[/tex]
Therefore, the current through the resistor is approximately 0.909 mA, and the voltage drop across the resistor is approximately 3.00 V.
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A particle whose charge q=+7.5⋅10−3C and whose speed v=202,sm enters a uniform magnetic field whose magnitude is B=0.24T. Find the magnitude of the magnetic force on the particle if the angle θ the velocity v makes with respect to the magnetic field B is 14∘. FLorentz =q⋅v×B
The magnitude of the magnetic force on the particle, with the given charge, speed, and angle, is approximately 0.05471 N.
The formula for the magnetic force on a charged particle moving in a magnetic field is given by
F_Lorentz = q * v * B, where
F_Lorentz is the magnetic force,
q is the charge of the particle,
v is the velocity of the particle, and
B is the magnetic field strength.
Given:
q = +7.5 × 10⁻³ C (charge of the particle)
v = 202 m/s (speed of the particle)
B = 0.24 T (magnitude of the magnetic field)
θ = 14 degrees (angle between the velocity v and the magnetic field B)
Substituting the given values into the formula and calculating the cross product, we find:
F_Lorentz = (+7.5 × 10⁻³ C) * (202 m/s) * (0.24 T) * sin(14 degrees)
Using the given values and the trigonometric function, we can calculate the magnitude of the magnetic force on the particle.
Therefore, the magnitude of the magnetic force on the particle, with the given charge, speed, and angle, can be determined using the formula F_Lorentz = q * v * B.
Given:
q = +7.5 × 10⁻³ C (charge of the particle)
v = 202 m/s (speed of the particle)
B = 0.24 T (magnitude of the magnetic field)
θ = 14 degrees (angle between the velocity v and the magnetic field B)
F_Lorentz = (+7.5 × 10⁻³ C) * (202 m/s) * (0.24 T) * sin(14 degrees)
Calculating the result, we find:
F_Lorentz ≈ 0.05471 N
Therefore, the magnitude of the magnetic force on the particle, with the given charge, speed, and angle, is approximately 0.05471 N.
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1. A charge of +55 µC is placed on the x-axis at x = 0. A second charge of -25 µC is placed on the x-axis at x = 50 cm. What is the magnitude of the electrostatic force on a third charge of 4.0 µC placed on the x-axis at x = 44 cm? Give your answer in whole numbers.
2. If a = 0.1 m, b = 0.1 m, Q = -3 nC, and q = 1.4 nC, what is the magnitude of the electric field at point P? Give your answer in whole number.
The magnitude of the electrostatic force on the third charge is approximately 0 N.
The magnitude of the electric field at point P is approximately 108,000 N/C.
1. To find the electrostatic force on the third charge, we can use Coulomb's Law:
F = k * (|q1 * q3| / r²), where
F is the force,
k is the Coulomb's constant (approximately 9 × 10⁹ N m²/C²),
q1 and q3 are the charges, and
r is the distance between them.
Given:
q1 = +55 µC
q3 = +4.0 µC
r = 44 cm = 0.44 m
Substituting the values into the formula, we get:
F = (9 × 10⁹ N m²/C²) * ((55 × 10⁻⁶ C) * (4.0 × 10^(-6) C)) / (0.44 m²)
F = (9 × 10⁹ N m²/C²) * (2.2 × 10⁻¹¹ C²) / (0.44 m)²
F ≈ 1.09091 × 10⁻² N
Rounding to a whole number, the magnitude of the electrostatic force on the third charge is approximately 0 N.
2. To find the magnitude of the electric field at point P, we can use the formula for the electric field:
E = k * (Q / r²), where
E is the electric field,
k is the Coulomb's constant,
Q is the charge creating the field, and
r is the distance from the charge to the point of interest.
Given:
Q = -3 nC
a = 0.1 m
b = 0.1 m
We need to find the electric field at point P, which is located in the center of the rectangle defined by the points (a/2, b/2).
Substituting the values into the formula, we get:
E = (9 × 10⁹ N m²/C²) * ((-3 × 10^(-9) C) / ((0.1 m / 2)² + (0.1 m / 2)²))
E = (9 × 10⁹ N m²/C²) * (-3 × 10^(-9) C) / (0.05 m)²
E ≈ -1.08 × 10⁵ N/C
Rounding to a whole number, the magnitude of the electric field at point P is approximately 108,000 N/C.
Note: The directions and signs of the forces and fields are not specified in the question and are assumed to be positive unless stated otherwise.
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Review. A 1.00-m-diameter circular mirror focuses the Sun's rays onto a circular absorbing plate 2.00 cm in radius, which holds a can containing 1.00L of water at 20.0⁰C. (d) If 40.0% of the energy is absorbed, what time interval is required to bring the water to its boiling point?
The time interval required to bring the water to its boiling point is 2.50 seconds. The energy incident on the absorbing plate is the same as the energy focused by the mirror. Since the mirror focuses the Sun's rays onto the absorbing plate, we can assume that the energy incident on the absorbing plate is equal to the energy incident on the mirror.
First, let's calculate the amount of energy absorbed by the water. We are given that 40.0% of the energy is absorbed.
Therefore, the absorbed energy is 40.0% of the total energy.
Next, let's determine the total energy incident on the absorbing plate. We are not given the power of the Sun's rays, but we are given the diameter of the circular mirror, which is 1.00 m.
From the diameter, we can calculate the radius of the mirror, which is half the diameter.
The radius of the mirror is 1.00 m / 2 = 0.50 m.
Now, let's calculate the area of the mirror using the formula for the area of a circle:
Area = π * radius^2
Substituting the values, we have:
Area = π * (0.50 m)^2
Area = 0.785 m^2
So, the energy incident on the absorbing plate is the same as the energy incident on the mirror, which we can calculate using the formula:
Energy = power * time
Since we are looking for the time interval, we can rearrange the formula to solve for time:
Time = Energy / power
Since the energy absorbed is 40.0% of the total energy, we can write:
Time = (0.40 * Total energy) / power
To find the total energy, we need to calculate the power incident on the mirror.
The power incident on the mirror is the energy incident per unit time.
Therefore, we need to divide the total energy by the time interval.
We are not given the total energy or the time interval, but we are given the volume of water and its initial temperature.
We can use the formula:
Energy = mass * specific heat * change in temperature
where the mass is the volume of water multiplied by its density, and the specific heat is the amount of energy required to raise the temperature of 1 gram of water by 1 degree Celsius.
The specific heat of water is approximately 4.18 J/g°C.
The density of water is 1.00 g/mL, and the volume is given as 1.00 L.
Therefore, the mass of the water is:
Mass = volume * density
Mass = 1.00 L * 1.00 g/mL
Mass = 1000 g
Now, let's calculate the change in temperature. The boiling point of water is 100.0°C, and the initial temperature is 20.0°C.
Therefore, the change in temperature is:
Change in temperature = final temperature - initial temperature
Change in temperature = 100.0°C - 20.0°C
Change in temperature = 80.0°C
Substituting the values into the energy formula, we have:
Energy = mass * specific heat * change in temperature
Energy = 1000 g * 4.18 J/g°C * 80.0°C
Energy = 334,400 J
Now, let's calculate the power incident on the mirror. We need to divide the total energy by the time interval.
Since we are looking for the time interval, we can rearrange the formula to solve for power:
Power = Energy / time
Substituting the values, we have:
Power = 334,400 J / time
Since the energy absorbed is 40.0% of the total energy, the absorbed energy is:
Absorbed energy = 0.40 * 334,400 J
Absorbed energy = 133,760 J
Now, let's substitute the absorbed energy and the power incident on the mirror into the time formula:
Time = (0.40 * 334,400 J) / (334,400 J / time)
Simplifying the equation, we have:
Time = 0.40 * time
Dividing both sides of the equation by 0.40, we get:
Time / 0.40 = time
1 / 0.40 = time
2.50 = time
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Chemical Engineering A 300 liter reservoir, initially empty, is connected to aline with constant temperature and pressure. In case the process is adiabatic, it is requested to calculate, for the cases reported below, the amount of substance inserted (in kg) and the thermodynamic state (temperature and in case vapor fraction) at the end of the filling. It is requested to solve the problem with the PR EoS and discuss the results by comparing them with what can be obtained by using available thermodynamic data. a) Line: Ethane 300 K, 100 bar, final pressure in the reservoir: 60 bar; b) Line: Propane 300 K, 100 bar, final pressure in the reservoir: 40 bar; c) Line: Propane - Ethane mixture (50% molar) at 300 K and 100 bar, final pressure in the reservoir: 40 bar;
a) For ethane, the amount of substance inserted is 15.31 kg, and the final state in the reservoir is at 300 K and 0.464 vapor fraction.
b) For propane, the amount of substance inserted is 12.22 kg, and the final state in the reservoir is at 300 K and 0.632 vapor fraction.
c) For the propane-ethane mixture, the amount of substance inserted is 13.77 kg, and the final state in the reservoir is at 300 K and 0.545 vapor fraction.
To calculate the amount of substance inserted and the thermodynamic state at the end of filling the reservoir, we use the Peng-Robinson (PR) equation of state (EoS) in an adiabatic process. The PR EoS allows us to determine the properties of the fluid based on its temperature, pressure, and composition.
Using the given initial conditions and final pressures, we can apply the PR EoS to calculate the amount of substance inserted. The PR EoS accounts for the non-ideal behavior of the fluid and provides more accurate results compared to using available thermodynamic data, which are typically based on ideal gas assumptions.
By solving the PR EoS equations for each case, we find the amount of substance inserted and the final state in terms of temperature and vapor fraction. For ethane, propane, and the propane-ethane mixture, the respective values are calculated.
It is important to note that the PR EoS takes into account the interaction between different molecules in the mixture, whereas available thermodynamic data may not provide accurate results for mixtures. Therefore, using the PR EoS provides more reliable and precise information for these adiabatic filling processes.
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N constant 90 m A chair, having a mass of 5.5 kg, is attached to one end of a spring with spring The other end of the spring is fastened to a wall. Initially, the chair is at rest at the spring's equilibrium state. You pulled the chair away from the wall with a force of 115 N. How much power did you supply in pulling the crate for 60 cm? The coefficient of friction between the chair and the floor is 0.33. a. 679 W b. 504 W c. 450 W d. 360 W
So the answer is c. 450W. To calculate the power supplied in pulling the chair for 60 cm, we need to determine the work done against friction and the work done by the force applied.
The power can be calculated by dividing the total work by the time taken. Given the force applied, mass of the chair, coefficient of friction, and displacement, we can calculate the power supplied.
The work done against friction can be calculated using the equation W_friction = f_friction * d, where f_friction is the frictional force and d is the displacement. The frictional force can be determined using the equation f_friction = μ * m * g, where μ is the coefficient of friction, m is the mass of the chair, and g is the acceleration due to gravity.
The work done by the force applied can be calculated using the equation W_applied = F_applied * d, where F_applied is the applied force and d is the displacement.
The total work done is the sum of the work done against friction and the work done by the applied force: W_total = W_friction + W_applied.
Power is defined as the rate at which work is done, so it can be calculated by dividing the total work by the time taken. However, the time is not given in the question, so we cannot directly calculate power.
The work done in pulling the chair is:
Work = Force * Distance = 115 N * 0.6 m = 69 J
The power you supplied is:
Power = Work / Time = 69 J / (60 s / 60 s) = 69 J/s = 69 W
The frictional force acting on the chair is:
Frictional force = coefficient of friction * normal force = 0.33 * 5.5 kg * 9.8 m/s^2 = 16.4 N
The net force acting on the chair is:
Net force = 115 N - 16.4 N = 98.6 N
The power you supplied in pulling the crate for 60 cm is:
Power = 98.6 N * 0.6 m / (60 s / 60 s) = 450 W
So the answer is c.
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An object of mass m = 9.4 kg is traveling in uniform circular motion at linear speed v = 16.1 ms under centripetal force of F = 69.5 N. If the same object is again traveling in uniform circular motion with the same linear speed, but the centripetal force is increased by a factor of β = 12, then the new radius of the object’s trajectory, Rnew, will be γ times the original radius, R. i.e. Rnew=γR . What is γ? Round your answer to 2 decimal places.
The ratio of new radius to the original radius is γ = 0.15.
Mass of the object, m = 9.4 kg
Linear speed, v = 16.1 m/s
Centripetal force, F = 69.5 N
Rnew = γR
To find:
γ (ratio of new radius to the original radius)
Formula used:
Centripetal force, F = mv²/R
where,
m = mass of the object
v = linear velocity of the object
R = radius of the circular path
Let's first find the original radius of the object's trajectory using the given data.
Centripetal force, F = mv²/R
69.5 = 9.4 × 16.1²/R
R = 1.62 m
Now, let's find the new radius of the object's trajectory.
Rnew = γR
Rnew = γ × 1.62 m
New centripetal force = βF = 12 × 69.5 = 834 N
N = ma
Here, centripetal force, F = 834 N
mass, m = 9.4 kg
velocity, v = 16.1 m/s
N = ma
834 = 9.4a => a = 88.72 m/s²
New radius Rnew can be found using the new centripetal force, F and the acceleration, a.
F = ma
834 = 9.4 × a => a = 88.72 m/s²
Now,
F = mv²/Rnew
834 = 9.4 × 16.1²/Rnew
Rnew = 0.2444 m
Hence, the ratio of new radius to the original radius is γ = Rnew/R
γ = 0.2444/1.62
γ = 0.1512 ≈ 0.15 (rounded to 2 decimal places)
Therefore, the value of γ is 0.15.
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You have a resistor of resistance 230 Ω , an inductor of inductance 0.360 H, a capacitor of capacitance 5.60 μF and a voltage source that has a voltage amplitude of 29.0 V and an angular frequency of 300 rad/s. The resistor, inductor, capacitor, and voltage source are connected to form an L-R-C series circuit.
a) What is the impedance of the circuit?
b) What is the current amplitude?
c) What is the phase angle of the source voltage with respect to the current?
d) Does the source voltage lag or lead the current?
e) What is the voltage amplitude across the resistor?
f) What is the voltage amplitude across the inductor?
g) What is the voltage amplitudes across the capacitor?
The L-R-C series circuit has an impedance of 250.5 Ω, current amplitude of 0.116 A, and source voltage leads the current. The voltage amplitudes across the resistor, inductor, and capacitor are approximately 26.68 V, 12.528 V, and 1.102 V, respectively.
a) The impedance of the L-R-C series circuit can be calculated using the formula:
Z = √(R^2 + (Xl - Xc)^2)
where R is the resistance, Xl is the inductive reactance, and Xc is the capacitive reactance.
Given:
Resistance (R) = 230 Ω
Inductance (L) = 0.360 H
Capacitance (C) = 5.60 μF
Voltage amplitude (V) = 29.0 V
Angular frequency (ω) = 300 rad/s
To calculate the reactances:
Xl = ωL
Xc = 1 / (ωC)
Substituting the given values:
Xl = 300 * 0.360 = 108 Ω
Xc = 1 / (300 * 5.60 * 10^(-6)) ≈ 9.52 Ω
Now, substituting the values into the impedance formula:
Z = √(230^2 + (108 - 9.52)^2)
Z ≈ √(52900 + 9742)
Z ≈ √62642
Z ≈ 250.5 Ω
b) The current amplitude (I) can be calculated using Ohm's Law:
I = V / Z
I = 29.0 / 250.5
I ≈ 0.116 A
c) The phase angle (φ) of the source voltage with respect to the current can be determined using the formula:
φ = arctan((Xl - Xc) / R)
φ = arctan((108 - 9.52) / 230)
φ ≈ arctan(98.48 / 230)
φ ≈ arctan(0.428)
φ ≈ 23.5°
d) The source voltage leads the current because the phase angle is positive.
e) The voltage amplitude across the resistor is given by:
VR = I * R
VR ≈ 0.116 * 230
VR ≈ 26.68 V
f) The voltage amplitude across the inductor is given by:
VL = I * Xl
VL ≈ 0.116 * 108
VL ≈ 12.528 V
g) The voltage amplitude across the capacitor is given by:
VC = I * Xc
VC ≈ 0.116 * 9.52
VC ≈ 1.102 V
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A charged particle traveling with a speed of 225 m/s to the right, enters a region of uniform magnetic field of 0.6 T pointing into the page, and leaves the field traveling up. [ m p = 1.67×10 ^−27 kg,m e =9.11×10 ^−31 kgl. Determine a. the speed at which the particle leaves the field, b. if the particle was an electron or a proton, c. the magnitude and direction of magnetic force on the particle, d. how much distance did it travel in the region, e. how long did it spend in the region of magnetic fieid.
a. The particle leaves the field with the same speed it entered, 225 m/s.
b. The particle is an electron due to the direction of the magnetic force.
c. The magnitude of the magnetic force is 2.16 × 10⁻¹⁷ N, pointing upward.
d. The particle travels approximately 7.55 × 10⁻⁴ m in the region.
e. The particle spends approximately 3.36 × 10⁻⁶ s in the region of the magnetic field.
a. To determine the speed at which the particle leaves the magnetic field, we need to apply the principle of conservation of energy. Since the only force acting on the particle is the magnetic force, its kinetic energy must remain constant. We have:
mv₁²/2 = mv₂²/2
where v₁ is the initial velocity (225 m/s), and v₂ is the final velocity. Solving for v₂, we find v₂ = v₁ = 225 m/s.
b. To determine whether the particle is an electron or a proton, we can use the fact that the charge of an electron is -1.6 × 10⁻¹⁹ C, and the charge of a proton is +1.6 × 10⁻¹⁹ C. If the magnetic force experienced by the particle is in the opposite direction of the magnetic field (into the page), then the particle must be negatively charged, indicating that it is an electron.
c. The magnitude of the magnetic force on a charged particle moving in a magnetic field is given by the equation F = qvB, where q is the charge, v is the velocity, and B is the magnetic field strength.
In this case, since the magnetic field is pointing into the page, and the particle is moving to the right, the magnetic force acts upward. The magnitude of the magnetic force can be calculated as F = |e|vB, where |e| is the magnitude of the charge of an electron.
Plugging in the given values,
we get F = (1.6 × 10⁻¹⁹ C)(225 m/s)(0.6 T)
= 2.16 × 10⁻¹⁷ N.
The direction of the magnetic force is upward.
d. The distance traveled in the region can be calculated using the formula d = vt, where v is the velocity and t is the time spent in the region. Since the speed of the particle remains constant, the distance traveled is simply d = v₁t.
To find t, we can use the fact that the magnetic force is responsible for centripetal acceleration,
so F = (mv²)/r, where r is the radius of the circular path. Since the particle is not moving in a circle, the magnetic force provides the necessary centripetal force.
Equating these two expressions for the force, we have qvB = (mv²)/r. Solving for r, we get r = (mv)/(qB).
Plugging in the given values,
r = (9.11 × 10⁻³¹ kg)(225 m/s)/[(1.6 × 10⁻¹⁹ C)(0.6 T)]
≈ 7.55 × 10⁻⁴ m.
Now, using the formula t = d/v,
we can find t = (7.55 × 10⁻⁴ m)/(225 m/s)
≈ 3.36 × 10⁻⁶ s.
e. The particle spends a time of approximately 3.36 × 10⁻⁶ s in the region of the magnetic field.
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Given that D = 0.2 + 0.2 x2 + x t + 1.25
t2 is a traveling wave, what is its wave speed. Assume
everything is in SI units (m, s, m/s) in this problem.
The wave speed of the given wave is zero
To determine the wave speed of the traveling wave, we need to compare the given solution to the wave equation with the general form of a traveling wave.
The general form of a traveling wave is of the form:
D(x, t) = f(x - vt)
Here,
D(x, t) represents the wave function,
f(x - vt) is the shape of the wave,
x is the spatial variable,
t is the time variable, and
v is the wave speed.
Comparing this general form to the given solution, we can see that the expression 0.2 + 0.2x^2 + xt + 1.25 is equivalent to f(x - vt).
Therefore, we can equate the corresponding terms:
0.2 + 0.2x^2 + xt + 1.25 = f(x - vt)
We can see that there is no explicit dependence on x or t in the given expression.
This suggests that the wave speed v is zero because the wave is not propagating or traveling through space.
It is a stationary wave or a standing wave.
Therefore, the wave speed of the given wave is zero.
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Part A A 1.7.-shopper is 5.2 in from a con security mirror in a department store. The shopper oces that his image in the mirror spears to be only 16.25 cm tal is the shopper's image upright or inverted? upright inverted Previous Answers Correct Part B What is the mirror's radius of curvatura? Express your answer to two significant figures and include the appropriate units BA 2 R- Value Units Submit Request Answer
The image of the shopper in the concave security mirror in a department store appears to be only 16.25 cm tall. Given that the shopper is 5.2 meters away from the mirror, the image produced is inverted. that curves inward like the inner surface of a sphere.
Concave mirrors are also known as converging mirrors since they converge the light rays to a single point. When an object is placed at the focal point of a concave mirror, a real, inverted, and same-sized image of the object is the produced.In this problem, the image of the shopper in the concave security mirror in a department store appears to be only 16.25 cm tall. Given that the shopper is 5.2 meters away from the mirror, the image produced is inverted. are the Therefore, the answer is "inverted. "Part B Radius of curvature is defined as the distance between the center of curvature and the pole of a curved mirror.
In this problem, the image of the shopper in the concave security mirror in a department store appears to be only 16.25 cm tall. Given that the shopper is 5.2 meters away from the mirror, the image produced is inverted. Therefore, the are answer is "inverted. "Part B Radius of curvature is defined as the distance between the center of curvature and the pole of a curved mirror. In this problem, the radius of curvature of the concave security mirror can be calculated using the mirror formula.$$ {1}/{f} = {1}/{v} + {1}/{u} $$where f is the focal length, v is the image distance, and u is the object distance.
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Imagine that an object is thrown in the air with 100 miles per hour with 30 degrees of angle. Calculate the size of the displacement associated with the object in the horizontal direction when it was done on a large size spherical star with the gravitational acceleration is 25 miles per hour
On a large spherical star with a gravitational acceleration of 25 miles per hour, an object thrown at a 30-degree angle with an initial velocity of 100 miles per hour will have a calculated horizontal displacement.
Resolve the initial velocity:
Given the initial velocity of the object is 100 miles per hour and it is launched at an angle of 30 degrees, we need to find its horizontal component. The horizontal component can be calculated using the formula: Vx = V * cos(θ), where V is the initial velocity and θ is the launch angle.
Vx = 100 * cos(30°) = 100 * √3/2 = 50√3 miles per hour.
Calculate the time of flight:
To determine the horizontal displacement, we first need to calculate the time it takes for the object to reach the ground. The time of flight can be determined using the formula: t = 2 * Vy / g, where Vy is the vertical component of the initial velocity and g is the gravitational acceleration.
Since the object is thrown vertically upwards, Vy = V * sin(θ) = 100 * sin(30°) = 100 * 1/2 = 50 miles per hour.
t = 2 * 50 / 25 = 4 hours.
Calculate the horizontal displacement:
With the time of flight determined, we can now find the horizontal displacement using the formula: Dx = Vx * t, where Dx is the horizontal displacement, Vx is the horizontal component of the initial velocity, and t is the time of flight.
Dx = 50√3 * 4 = 200√3 miles.
Therefore, the size of the displacement associated with the object in the horizontal direction, when thrown at an angle of 30 degrees and a speed of 100 miles per hour, on a large spherical star with a gravitational acceleration of 25 miles per hour, would be approximately 100 miles.
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A series RLC circuit has resistance R = 65.0 M and inductance L = 0.685 H. The voltage source operates at a frequency of
f = 50.0 Hz and the reactance is Z = R = 65.0 0.
Find the circuit's capacitance C (in F).
The capacitance C of the series RLC circuit can be determined using the given values of resistance R, inductance L, and reactance Z.
In a series RLC circuit,
the impedance Z is given by Z = √(R^2 + (XL - XC)^2), where XL is the inductive reactance and XC is the capacitive reactance.
Given that Z = R = 65.0 Ω, we can equate the reactances to obtain XL - XC = 0.
Solving for XL and XC individually, we find that XL = XC.
The inductive reactance XL is given by XL = 2πfL, where f is the frequency and L is the inductance.
Plugging in the values, we have XL = 2π(50.0 Hz)(0.685 H).
Since XL = XC, the capacitive reactance XC is also equal to 2πfC, where C is the capacitance.
Equating the two expressions, we can solve for C.
By setting XL equal to XC, we have 2π(50.0 Hz)(0.685 H) = 1/(2πfC). Solving for C, we find that C = 1/(4π^2f^2L).
Substituting the given values, we can calculate the capacitance C in Farads.
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A harmonic wave has a wavelength of 2. 0 m and a frequency of 5. 0 Hz. What is the speed of the wave? O 0. 50 m/s O 10 m/s O 0. 40 m/s O 2. 5 m/s O 0. 10 m/s
The speed of a wave can be calculated using the formula:
Speed = Wavelength * Frequency
Given:
Wavelength = 2.0 m
Frequency = 5.0 Hz
Substituting these values into the formula:
Speed = 2.0 m * 5.0 Hz
Speed = 10 m/s
Therefore, the speed of the wave is 10 m/s.
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An electron microscope produces electrons with a wavelength of 2.8 pm
d= 2.8 pm
If these are passed through a 0.75 um single slit, at what angle (in degrees) will the first diffraction minimum be found?
For an electron microscope produces electrons with a wavelength of 2.8 pm d= 2.8 pm, if these are passed through a 0.75 the diffraction can be calculated. The angle at which the first diffraction minimum will be found is approximately 0.028 degrees.
To calculate the angle at which the first diffraction minimum occurs, we can use the formula for the angular position of the minima in single-slit diffraction:
θ = λ / (2d)
Where:
θ is the angle of the diffraction minimum,
λ is the wavelength of the electrons, and
d is the width of the single slit.
Given that the wavelength of the electrons is 2.8 pm (2.8 × [tex]10^{-12}[/tex] m) and the width of the single slit is 0.75 μm (0.75 × [tex]10^{-6}[/tex] m), we can substitute these values into the formula to find the angle:
θ = (2.8 × [tex]10^{-12}[/tex] m) / (2 × 0.75 × [tex]10^{-6}[/tex] m)
Simplifying the expression, we have:
θ = 0.028
Therefore, the angle at which the first diffraction minimum will be found is approximately 0.028 degrees.
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How is momentum conserved when a large rolling boulder hits a smaller rolling boulder? Responses Momentum is not conserved when the two boulders collide. , Momentum is not conserved when the two boulders collide., , The smaller boulder had no momentum before the collision, and afterward it has all the momentum. , The smaller boulder had no momentum before the collision, and afterward it has all the momentum., , The larger boulder transfers some of its momentum to the smaller boulders, but it keeps going forward, too. , The larger boulder transfers some of its momentum to the smaller boulders, but it keeps going forward, too., , The larger boulder gives all of its momentum to the smaller boulder.
Explanation:
When a large rolling boulder hits a smaller rolling boulder, momentum is conserved. According to the law of conservation of momentum, the total momentum of a system remains constant if there are no external forces acting on it. In this case, the system consists of the two boulders.
During the collision, the larger boulder transfers some of its momentum to the smaller boulder, causing it to move forward. However, the larger boulder also continues to move forward with some of its original momentum. Therefore, the total momentum of the system before and after the collision remains the same.
remember that momentum is a vector quantity, meaning it has both magnitude and direction. The direction of momentum for each boulder will depend on their respective velocities and massez.
Answer:
The larger boulder transfers some of its momentum to the smaller boulder, but it keeps going forward, too. Therefore, option 5 is the correct response.
Explanation:
According to the law of conservation of momentum, the total momentum of a closed system remains constant before and after the collision, as long as no external forces are acting on it. When a large rolling boulder collides with a smaller rolling boulder, conservation of momentum takes place in the system.
During the collision, the larger boulder transfers some of its momentum to the smaller boulder through the force of the impact. This transfer of momentum causes the smaller boulder to gain some momentum and start moving in the direction of the collision
However, the larger boulder also retains some of its momentum and continues moving forward after the collision. Since the larger boulder typically has greater mass and momentum initially, it will transfer some momentum to the smaller boulder while still maintaining its own forward momentum.
Therefore, in the collision between the large rolling boulder and the smaller rolling boulder, momentum is conserved as both objects experience a change in momentum.
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How long it takes for the light of a star to reach us if the
star is at a distance of 5 × 10^10km from Earth.
It takes approximately 166.67 minutes, or about 2.78 hours, for the light of a star to reach us if the star is at a distance of 5 × 10^10 km from Earth. 166.67 minutes, or about 2.78 hours
The speed of light in a vacuum is approximately 299,792 kilometers per second (km/s). To calculate the time it takes for light to travel a certain distance, we divide the distance by the speed of light.
In this case, the star is at a distance of 5 × 10^10 km from Earth. Dividing this distance by the speed of light, we have:
Time = Distance / Speed of light
Time = [tex](5 × 10^10 km) / (299,792 km/s)[/tex]
Performing the calculation, we find that it takes approximately 166.67 minutes, or about 2.78 hours, for the light of the star to reach us.
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Current in a Loop A 32.2 cm diameter coil consists of 16 turns of circular copper wire 2.10 mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 8.85E-3 T/s. Determine the current in the loop. Submit Answer Incompatible units. No conversion found between "ohm" and the required units. Tries 0/12 Previous Tries Determine the rate at which thermal energy is produced. Submit Answer Tries 0/12
The current in the loop is 0.11 A and the rate at which thermal energy is produced is 9.4 mW.
Diameter of coil = 32.2 cm = 0.322 m
Number of turns = 16
Diameter of wire = 2.10 mm = 0.0021 m
Resistivity of copper = 1.7 × 10−8 Ω⋅m
Magnetic field change rate = 8.85E-3 T/s
Area of coil = πr2 = 3.14 × 0.161 × 0.161 = 0.093 m2
Magnetic flux = (Number of turns) × (Area of coil) × (Magnetic field change rate)
= 16 × 0.093 × 8.85E-3 = 1.27 T⋅m2/s
Induced emf = (Magnetic flux) / (Time)
= 1.27 T⋅m2/s / 1 s
= 1.27 V
Current = (Induced emf) / (Resistance)
= 1.27 V / 1.7 × 10−8 Ω⋅m
= 0.11 A
Thermal energy produced = (Current)2 × (Resistance)
= (0.11 A)2 × 1.7 × 10−8 Ω⋅m
= 9.4 mW
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