(a) The angle between the normal to the surface of the mirror and due south should be approximately 0.994 degrees.
(b) If the mirror is misaligned by an additional 0.0030 degrees, the reflected ray will miss the detector by approximately 0.035 meters due east.
(a) To determine the angle between the normal to the mirror's surface and due south, we can use trigonometry. Since the mirror is due north of the laser and the detector is due east, the angle formed by the normal to the mirror and due south is the complement of the angle formed by the laser, mirror, and detector. Using the right triangle formed by the laser, mirror, and detector, we can find that the angle is approximately 0.994 degrees.
(b) If the mirror is misaligned by an additional 0.0030 degrees, the reflected ray will deviate from the correct angle. To calculate the horizontal distance by which the reflected ray misses the detector, we can use trigonometry again. The horizontal distance is equal to the vertical distance (125 m) multiplied by the tangent of the misalignment angle. Therefore, the reflected ray will miss the detector by approximately 0.035 meters due east.
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What is the ratio of the orbital velocity of a terrestrial planet orbiting at 5.00 AU from its star to that of a giant planet orbiting at 19.00 AU? NOTE: You may assume circular orbits.
The ratio of the velocities of the two planets is 0.482
The velocity v of the planet in its circular orbit can be calculated using the following formula: v = 2πr/T
So, v = 2πr/ T
Where T is the time period of revolution of the planet around the star, r is the radius of the circular orbit, and v is the velocity of the planet in its orbit.
Since we can assume circular orbits, we have v ∝ r^-1/2.
Therefore, the ratio of velocities of the two planets is given as follows:
V1 / V2 = (r1 / r2)^(1/2)
Where, V1 is the velocity of the terrestrial planet orbiting at 5.00 AU, r1 is the radius of the orbit of the terrestrial planet around the star, V2 is the velocity of the giant planet orbiting at 19.00 AU, and r2 is the radius of the orbit of the giant planet around the star.
Therefore, substituting r1 = 5.00 AU, r2 = 19.00 AU, we get the ratio of the velocities of the two planets as:
V1 / V2 = (5/19)^(1/2)
V1 / V2 = 0.482
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Microscope used for detecting defects in electrical conductors and computer chips is (A) Atomic force microscope (B) Scanning tunnelling microscope (C) X.ray crystallography (D) Scanning electron microscope.
Microscope used for detecting defects in electrical conductors and computer chips is (D) Scanning electron microscope.
A scanning electron microscope (SEM) is commonly used for detecting defects in electrical conductors and computer chips.
SEM uses a focused beam of electrons to scan the surface of a sample and create a highly magnified image. It provides detailed information about the surface morphology, composition, and defects of the sample.
SEM is particularly useful in semiconductor industry and materials science for analyzing microelectronics and identifying defects in electrical conductors and computer chips.
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An electric stove burner of radius 40 cm is at a temperature of 300 °C. If a = 5.67 x 10-8 W/m².K4, at what rate is the burner radiating energy? Assume the emissivity & = 0.52 8000 W 320 W O 1600 W 0 4W O 80 W
The rate of the burner radiating energy is 9.10937086 × 1025 m10 kg / s3.
To calculate the rate at which the burner is radiating energy, we can use the Stefan-Boltzmann law, which states that the power radiated per unit area by an object is proportional to the fourth power of its temperature and is given by:
P = εσAΔT⁴
where P is the power radiated, ε is the emissivity, σ is the Stefan-Boltzmann constant (σ = 5.67 x 10^-8 W/m².K^4), A is the surface area of the burner, and ΔT is the temperature difference between the burner and its surroundings.
Given:
Radius of the burner (r) = 40 cm = 0.4 m
Temperature of the burner (T) = 300 °C = 573 K
Emissivity (ε) = 0.52
Stefan-Boltzmann constant (σ) = 5.67 x 10^-8 W/m².K^4
First, we need to calculate the surface area of the burner:
A = πr²
Substituting the values:
A = π(0.4 m)²
Now, we can calculate the power radiated by the burner:
P = εσAΔT⁴
P = (0.52)(5.67 x 10^-8 W/m².K^4)(π(0.4 m)²)(573 K - 293 K)⁴
= 9.10937086 × 1025 m10 kg / s3
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An image formed by a convex mirror (f=-31.3 cm) has a magnification of 0.186. How much should the object be moved to double the size of the image? (Give the displacement with a sign that indicates the direction. Assume that the displacement toward the mirror is positive) Number Units cm Question 3 Numeric Fill in the Units Question 4
The object should be moved by approximately 0.01595 cm toward the mirror to double the size of the image.
To solve this problem, we can use the mirror formula and the magnification formula for convex mirrors. The mirror formula is given by:
1/f = 1/v - 1/u
Where:
f is the focal length of the mirror,
v is the image distance (distance of the image from the mirror), and
u is the object distance (distance of the object from the mirror).
The magnification formula is given by:
m = -v/u
Where:
m is the magnification.
We are given the focal length f = -31.3 cm and the magnification m = 0.186. We need to find the displacement of the object (u) that will result in doubling the size of the image.
Let's assume the initial object distance is u, and the initial image distance is v.
According to the magnification formula, we have:
m = -v/u
0.186 = -v/u
Rearranging the equation, we find:
v/u = -0.186
Now, let's consider the new object distance (u') that will result in doubling the size of the image. Since the image size is doubled, the new magnification (m') will be 2 times the initial magnification:
m' = 2m
m' = 2(0.186)
m' = 0.372
Using the magnification formula, we have:
m' = -v'/u'
0.372 = -v'/u'
Rearranging the equation, we find:
v'/u' = -0.372
Comparing this equation with the previous equation (v/u = -0.186), we can see that the ratio of v' to u' is double the ratio of v to u.
Therefore, to double the size of the image, we need to double the ratio v/u. Since v/u = -0.186, we can find the new ratio by multiplying -0.186 by 2:
v'/u' = -0.186 × 2
v'/u' = -0.372
Now, let's use the mirror formula to find the new image distance (v') in terms of u':
1/f = 1/v' - 1/u'
Substituting the given value of f = -31.3 cm and the ratio v'/u' = -0.372, we can solve for v':
1/(-31.3) = 1/v' - 1/u'
-0.0319 = 1/v' - 1/u'
Since we are interested in finding the displacement of the object, let's rearrange the equation to solve for u':
1/u' = 1/v' + 0.0319
1/u' = -1/u' + 0.0319
2/u' = 0.0319
u' = 0.0319/2
u' = 0.01595 cm
Therefore, the sign indicating the direction is positive, which means the object should be displaced towards the mirror.
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Two side-by-side loudspeakers at the origin emit 684 Hz sound waves on a day when the speed of sound is 342 m/s. A microphone 1.00 m away on the x-axis detects a maximum in the sound intensity. Then one of the speakers is moved slowly along the y-axis.
How far does it move before the microphone first detects a minimum in the sound intensity?
Express your answer with the appropriate units.
The speaker needs to move approximately 0.0855 meters (or 8.55 cm) along the y-axis before the microphone first detects a minimum in the sound intensity. When two loudspeakers emit sound waves, they create interference patterns due to the superposition of the waves. Depending on the relative phases of the waves, constructive or destructive interference can occur, resulting in areas of maximum or minimum sound intensity.
In this scenario, the microphone detects a maximum in sound intensity when it is located at a distance of 1.00 meter on the x-axis. As one of the speakers is moved along the y-axis, the phase difference between the waves reaching the microphone changes.
To determine the distance the speaker needs to move for the microphone to detect a minimum, we can consider the condition for destructive interference. Destructive interference occurs when the path difference between the two waves is equal to half a wavelength.
The wavelength of the sound wave can be calculated using the formula:
λ = v/f,
where:
λ is the wavelength,
v is the speed of sound (342 m/s),
f is the frequency of the sound wave (684 Hz).
λ = 342 m/s / 684 Hz,
λ = 0.5 m.
Since the microphone is initially at a maximum, the path difference between the two waves is equal to an odd number of half wavelengths (1/2, 3/2, 5/2, etc.) for destructive interference to occur.
The path difference (d) between the two waves can be determined using the Pythagorean theorem:
d = [tex]\sqrt(x^2 + y^2),[/tex]
where:
x is the initial distance of the microphone on the x-axis (1.00 m),
y is the distance the speaker is moved along the y-axis (unknown).
For destructive interference, the path difference (d) should be equal to an odd number of half wavelengths. Therefore, we have:
d = (2n + 1) * (λ/2),
where n is an integer.
Substituting the values, we get:
[tex]\sqrt(1.00^2 + y^2)[/tex] = (2n + 1) * (0.5/2).
Simplifying the equation, we have:
[tex]\sqrt(1 + y^2)[/tex] = (2n + 1) * 0.25.
Squaring both sides, we obtain:
[tex]1 + y^2 = (2n + 1)^2[/tex]* 0.0625.
[tex]1 + y^2 = (4n^2 + 4n + 1)[/tex]* 0.0625.
Simplifying further:
[tex]1 + y^2 = n^2[/tex] + n + 0.25.
Rearranging the equation, we have:
y^2 = n^2 + n - 0.75.
To find the smallest positive value of y that satisfies the equation, we substitute n = 0 and solve for y:
[tex]y^2 = 0^2[/tex] + 0 - 0.75,
[tex]y^2 = -0.75.[/tex]
Since we are looking for a real positive value of y, we discard this solution.
Next, we substitute n = 1 and solve for y:
[tex]y^2 = 1^2[/tex]+ 1 - 0.75,
[tex]y^2[/tex] = 1.25,
y ≈ ± 1.12.
Since we are interested in the positive value of y, we take y ≈ 1.12 m.
Therefore, the speaker needs to move approximately 1.12 meters along the y-axis for the microphone to first detect a minimum in the sound intensity.
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Consider the Continuous-Time LTI system as follows: dy(t) - + 2y(t) = x(t) dt Analyse the output y(t) if x(t) = e-tu(t) using Fourier Transform
The output of the Continuous-Time LTI system with input x(t) = e^(-t)u(t), analyzed using Fourier Transform, is given by y(t) = (1/(s+2)) * (1/(s+1)), where s is the Laplace variable.
To analyze the output of the given Continuous-Time LTI system, we can use the Laplace transform, which is the continuous-time counterpart of the Fourier transform. The Laplace transform of the input signal x(t) = e^(-t)u(t) can be obtained as X(s) = 1/(s+1), where s is the Laplace variable.
Applying the Laplace transform to the differential equation describing the system, we have (sY(s) - y(0)) + 2Y(s) = X(s), where Y(s) is the Laplace transform of the output signal y(t) and y(0) represents the initial condition of y(t) at t=0. Since the system is assumed to be at rest initially, y(0) is zero.
Rearranging the equation and substituting X(s), we get Y(s) = 1/[(s+2)(s+1)]. This is the Laplace transform of the output signal y(t).
To find the time-domain expression for y(t), we need to inverse Laplace transform Y(s). The inverse Laplace transform of 1/[(s+2)(s+1)] can be computed using partial fraction decomposition. By decomposing the expression, we obtain Y(s) = (1/(s+2)) * (1/(s+1)).
Finally, taking the inverse Laplace transform of Y(s), we find the output signal y(t) = e^(-2t)u(t) - e^(-t)u(t).
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A toy chest and its contents have a combined weight of W = 220 N. The coefficient of static friction between toy chest and floor μs is 0.460. The child in the figure attempts to move the chest across the floor by pulling on an attached rope. (a) If θ is 42.0°, what is the magnitude of the force F⎯⎯⎯ that the child must exert on the rope to put the chest on the verge of moving? Determine (b) the value of θ for which F is a minimum and (c) that minimum magnitude.
(a) the magnitude of the force the child must exert on the rope to put the chest on the verge of moving is 101.2 N. (b) the minimum force will be when θ = 0°. (c) The minimum magnitude of the force (F) required to move the chest when θ = 0° is 220 N.
We'll break down the forces acting on the toy chest and use the conditions for static equilibrium.
Weight of the toy chest and its contents (W) = 220 N
Coefficient of static friction (μs) = 0.460
Angle (θ) = 42.0°
(a) To put the chest on the verge of moving, the force applied by the child on the rope (F⎯⎯⎯) must overcome the static friction.
The force of static friction (f_s) can be calculated using the formula:
f_s = μs * N
The normal force (N) is equal to the weight of the chest (W):
N = W
So, the force of static friction is:
f_s = μs * W
The force applied by the child on the rope (F⎯⎯⎯) must be equal to the force of static friction for the chest to start moving. Therefore:
F⎯⎯⎯ = f_s
Substituting the values:
F⎯⎯⎯ = μs * W
F⎯⎯⎯ = 0.460 * 220 N
Calculate F⎯⎯⎯:
F⎯⎯⎯ = 101.2 N
Therefore, the magnitude of the force the child must exert on the rope to put the chest on the verge of moving is 101.2 N.
(b) To find the value of θ for which F is a minimum, we need to consider the angle at which the force required to move the chest is minimized. This occurs when the angle of the applied force is parallel to the direction of motion (parallel to the floor).
So, the minimum force will be when θ = 0°.
(c) At θ = 0°, the minimum magnitude of the force can be found by considering only the horizontal component of the weight (W) acting in the direction of motion.
The horizontal component of the weight is:
W_horizontal = W * cos(θ)
Substituting the values:
W_horizontal = 220 N * cos(0°)
Calculate W_horizontal:
W_horizontal = 220 N * 1
W_horizontal = 220 N
Therefore, the minimum magnitude of the force (F) required to move the chest when θ = 0° is 220 N.
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A 25-turn circular loop of wire is placed into a magnetic field with initial magnitude 5.3 T. The magnetic field is perpendicular to the surface of the loop. Over a period of 0.15 seconds, the strength of the field is decreased to 1.9 T and as the field decreases a 5.5 V emf is induced in the loop. Calculate the diameter of the loop of wire.
The diameter of the loop of wire is approximately 71.6 cm.
The electromotive force (emf) induced in a loop of wire is given by Faraday's law of electromagnetic induction, which states that the emf is equal to the rate of change of magnetic flux through the loop. In this case, the emf is given as 5.5 V.
The change in magnetic field is from an initial magnitude of 5.3 T to a final magnitude of 1.9 T. The rate of change of the magnetic field can be calculated by dividing the difference in field strengths by the time taken. Therefore, the rate of change of the magnetic field is (5.3 T - 1.9 T) / 0.15 s = 21.33 T/s.
Since the loop is circular, the magnetic flux through the loop is equal to the product of the magnetic field strength, the area of the loop, and the number of turns. Therefore, we can write the equation as emf = -N(dB/dt)A, where N is the number of turns and A is the area of the loop.
In this case, we are given that the number of turns is 25. Let's assume the diameter of the loop is D. The area of the loop can be calculated using the formula A = π(D/2)^2 = πD^2/4.
Substituting the given values into the equation, we have 5.5 V = -25(21.33 T/s)(πD^2/4). Simplifying, we can solve for the diameter D:
D^2 = -4(5.5 V) / (-25(21.33 T/s)(π/4)) = 0.512
Taking the square root of both sides, we find D = 0.716 meters or 71.6 centimeters. Therefore, the diameter of the loop of wire is approximately 71.6 cm.
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Part A What is the maximum speed with which a 1200-kg car can round a turn of radius 85.0 m on a flat road if the coefficient of static friction between tires and road is 0.40? Express your answer to two significant figures and include the appropriate units. μA ? Um Value Units Submit Request Answer Part B Is this result independent of the mass of the car? yes Request Answer O no Submit
Part A: The maximum speed with which the car can round the turn is approximately 14.7 m/s.
Part B: No, the result is not independent of the mass of the car. The maximum speed is directly affected by the mass of the car.
Part A:
To calculate the maximum speed with which the car can round the turn, we can use the centripetal force equation:
F = m * (v^2 / r)
Where:
F is the maximum static frictional force (provided by the coefficient of static friction)
m is the mass of the car
v is the velocity of the car
r is the radius of the turn
We can rearrange the equation to solve for v:
v = sqrt((F * r) / m)
The maximum static frictional force can be calculated as:
F = μ * N
Where:
μ is the coefficient of static friction
N is the normal force, which is equal to the weight of the car (mg)
Substituting the value of F into the previous equation, we have:
v = sqrt((μ * N * r) / m)
Now we can calculate the maximum speed:
μ = 0.40 (coefficient of static friction)
m = 1200 kg (mass of the car)
r = 85.0 m (radius of the turn)
g = 9.8 m/s^2 (acceleration due to gravity)
N = mg = 1200 kg * 9.8 m/s^2 = 11760 N
v = sqrt((0.40 * 11760 N * 85.0 m) / 1200 kg) ≈ 14.7 m/s
Part B:
No, the result is not independent of the mass of the car. The maximum speed is directly affected by the mass of the car. A car with a larger mass will require a greater centripetal force to round the turn at the same radius, which means it will have a lower maximum speed.
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What is the ideal banking angle (in degrees) for a turn of 1.1 km radius for a car traveling 89 km/h?
The ideal banking angle for a turn of 1.1 km radius with a car traveling at 89 km/h is approximately 3.19 degrees.
To find the ideal banking angle for a turn, we can use the equation:
tan(θ) = [tex](v^2) / (g * r)[/tex]
where θ is the banking angle, v is the velocity of the car, g is the acceleration due to gravity (approximately[tex]9.8 m/s^2[/tex]), and r is the radius of the turn.
First, let's convert the given values to the appropriate units:
v = 89 km/h = 24.72 m/s
r = 1.1 km = 1100 m
g = [tex]9.8 m/s^2[/tex]
Plugging in these values into the equation, we have:
tan(θ) =[tex](24.72^2) / (9.8 * 1100)[/tex]
Calculating the numerator:
[tex]24.72^2 = 611.5024[/tex]
Calculating the denominator:
9.8 * 1100 = 10780
Substituting the values back into the equation:
tan(θ) = 611.5024 / 10780
Using inverse tangent to find the angle:
θ = atan(611.5024 / 10780)
Calculating the angle in degrees:
θ ≈ 3.19 degrees
Therefore, the ideal banking angle for a turn of 1.1 km radius with a car traveling at 89 km/h is approximately 3.19 degrees.
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A car's bumper is designed to withstand a 4.32-km/h (1.2-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.240 m while bringing a 850-kg car to rest from an initial speed of 1.2 m/s.
The magnitude of the average force on the bumper is approximately 1700 Newtons.
To calculate the magnitude of the average force on the bumper, we can use the work-energy principle. The work done on an object is equal to the change in its kinetic energy.
The work done on the car is given by the product of the average force on the bumper and the distance over which the force is applied. This work is equal to the change in kinetic energy of the car.
The initial kinetic energy of the car is (1/2) * mass * (initial velocity)^2, and the final kinetic energy is zero since the car comes to rest. Therefore, the change in kinetic energy is equal to the initial kinetic energy.
The work done on the car is also equal to the force on the bumper times the distance over which it acts:
Force * distance = (1/2) * mass * (initial velocity)^2
We can rearrange this equation to solve for the force:
Force = (1/2) * mass * (initial velocity)^2 / distance
Plugging in the given values:
mass = 850 kg
initial velocity = 1.2 m/s
distance = 0.240 m
Force = (1/2) * 850 kg * (1.2 m/s)^2 / 0.240 m
Force ≈ 1700 N
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A hockey puck slides off the edge of a platform with an initial velocity of 25.9 m/s horizontally. The height of the platform above the ground is 7.5 m. What is the magnitude of the velocity (m/s) of the puck just before it touches the ground? Ignore air resistance, and take DOWNWARD AS POSITIVE.
The magnitude of the velocity of the puck just before it touches the ground is approximately 31.7 m/s.
To determine the velocity of the puck just before it touches the ground, we can use the principle of conservation of energy. Initially, the puck only has gravitational potential energy, which is converted to kinetic energy as it falls.
The gravitational potential energy of the puck at the height of 7.5 m can be calculated using the formula PE = mgh, where m is the mass of the puck, g is the acceleration due to gravity, and h is the height.
Since the mass of the puck is not provided, we can cancel it out when comparing the initial and final energies. Initially, the puck only has gravitational potential energy, so PE = mgh. When the puck touches the ground, it has both kinetic energy and no potential energy, so KE = 1/2 mv^2.
Using the conservation of energy, we can equate the initial potential energy to the final kinetic energy: mgh = 1/2 mv^2. We can cancel out the mass and rearrange the equation to solve for v: v = √(2gh). Plugging in the values of g = 9.8 m/s² and h = 7.5 m, we find v ≈ 31.7 m/s.
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In the Challenger Deep of the Marianas Trench, the depth of seawater is 10.9 km and the pressure is 1.16 x 108 Pa (about 1.15 x 10³ atm). (a) If a cubic meter of water is taken from the sur- face to this depth, what is the change in its volume? (Normal atmos- pheric pressure is about 1.0 X 10' Pa. Assume that k for seawater is the same as the freshwater value given in Table 11.2.) (b) What is the density of seawater at this depth? (At the surface, seawater has a density of 1.03 x 10³ kg/m³.) Bulk modulus of seawater is 2.2X10⁹ Pa.
a) The change in volume of water when taken from the surface to the Challenger Deep is approximately -1159 m³. This means that the volume decreases.
b) The density of seawater at the depth of the Challenger Deep is approximately 976.019 kg/m³.
(a) To calculate the change in volume of water when taken from the surface to the depth of the Challenger Deep, we can use the equation:
ΔV = V0 * (1 - (P1 / P0))
where ΔV is the change in volume, V0 is the initial volume (1 m³), P1 is the pressure at the depth (1.16 x 10⁸ Pa), and P0 is the initial pressure (1.0 x 10⁵ Pa).
Substituting the values into the equation, we have:
ΔV = 1 * (1 - (1.16 x 10⁸ / 1.0 x 10⁵))
= 1 * (1 - 1160)
≈ 1 * (-1159)
(b) The density of seawater at the depth can be calculated using the equation:
ρ = ρ0 * (1 - (P1 / K))
where ρ is the density at the depth, ρ0 is the initial density (1.03 x 10³ kg/m³), P1 is the pressure at the depth (1.16 x 10⁸ Pa), and K is the bulk modulus of seawater (2.2 x 10⁹ Pa).
Substituting the values into the equation, we have:
ρ = 1.03 x 10³ * (1 - (1.16 x 10⁸ / 2.2 x 10⁹))
≈ 1.03 x 10³ * (1 - 0.0527)
≈ 1.03 x 10³ * 0.9473
≈ 976.019 kg/m³
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An air-filled toroidal solenoid has a mean radius of 15.5 cm and a cross-sectional area of 5.05 cm^2. When the current is 12.5 A, the energy stored is 0.385 J
How many turns does the winding have?
An air-filled toroidal solenoid has a mean radius of 15.5 cm and a cross-sectional area of 5.05 [tex]cm^2[/tex]. When the current is 12.5 A, the energy stored is 0.385 J. So it have 102 turns.
The energy stored in an inductor (solenoid) can be calculated using the formula: E = [tex](1/2) * L * I^2[/tex]
where E is the energy stored, L is the inductance, and I is the current.
Given that the energy stored is 0.385 J and the current is 12.5 A, we can rearrange the formula to solve for the inductance: L = [tex](2 * E) / I^2[/tex]
Substituting the given values, we have:
L = [tex](2 * 0.385 J) / (12.5 A)^2[/tex]
L = 0.00616 H
The inductance of a toroidal solenoid is given by the formula:
L = (μ₀ * [tex]n^2[/tex] * A) / (2π * R)
where μ₀ is the permeability of free space, n is the number of turns, A is the cross-sectional area, and R is the mean radius.
Rearranging the formula, we can solve for n:
n = √[(2 * π * R * L) / (μ₀ * A)]
Substituting the given values and constants, we find:
n = √[(2 * π * 0.155 m * 0.00616 H) / (4π * [tex]10^-7[/tex]T * m/A * 0.0505 [tex]m^2[/tex])]
n ≈ 102 turns
Therefore, the winding of the air-filled toroidal solenoid has 102 turns.
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The two following waves travel along the same string y₁ = (2.1 m) sin [2rx - 20rt)] 32 = (2.1 m) sin [2nx - 20nt + ] where y₁, 32, and z are in meters and t is in seconds. A traveling wave is produced when the two waves are combined. Show the steps (a) Find the equation of the combined wave (10 pt) (b) What is the transverse speed of a particle on the string at position z = m when t=second?
(a) Combined wave equation: y = y₁ + y₂. (b) Transverse speed at z = m, t = second: Differentiate combined wave equation with respect to time, evaluate at given position and time.
(a) To find the equation of the combined wave, we can simply add the two individual waves:
y = y₁ + y₂
y = (2.1 m) sin [2rx - 20rt)] + (2.1 m) sin [2nx - 20nt + Φ]
Using the trigonometric identity sin (A + B) = sin A cos B + cos A sin B, we can simplify this expression:
y = (2.1 m) [sin (2rx - 20rt) cos Φ + cos (2rx - 20rt) sin Φ + sin (2nx - 20nt) ]
We can further simplify this expression by defining the wave number difference Δk = 2n - 2r and the phase difference ΔΦ = Φ. Then, the combined wave can be written as:
y = (2.1 m) [sin (2rx - 20rt + ΔΦ/2) cos (Δkx/2) ]
(b) The transverse speed of a particle on the string at position z = m when t=second can be found by taking the derivative of the combined wave equation with respect to time:
v = ∂y/∂t = - (2.1 m) (20r/Δk) cos (Δkx/2) sin (2rx - 20rt + ΔΦ/2)
Substituting the given values, we get:
v = - (2.1 m) (20r/(2n - 2r)) cos [(2n - 2r)x/4] sin [2rx - 20rt + Φ/2]
At z = 0.5 m and t = 1 second, the transverse speed of a particle on the string is:
v = - (2.1 m) (20(2π)(2)/(2π)(2) ) cos [(2π)(2)/4] sin [2(2π)(0.5) - 20(1) + 0]/2
v = -7.5 m/s
Therefore, the transverse speed of the particle at z = 0.5 m and t = 1 second is -7.5 m/s.
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Suppose the length of a clock's pendulum is increased by 1,300%, exactly at noon one day. What time will it read 24.00 hours later, assuming it the pendulum has kept perfect time before the change? Perform the colculation to at least five-digit precision.
The time on the clock will read approximately 86.5333 hours after the pendulum length is increased.
To calculate the new time on the clock after 24 hours with the increased pendulum length, we need to consider the relationship between the period of the pendulum and the time it takes for one complete swing.
The period of a pendulum is given by the equation:
T = 2π√(L/g)
Where:
T = Period of the pendulum
L = Length of the pendulum
g = Acceleration due to gravity (approximately 9.8 m/s^2)
Let's assume the original length of the pendulum is L0, and the new length after the increase is L1 = 13 * L0.
The ratio of the periods of the pendulum with the new and original lengths can be expressed as:
T1 / T0 = √(L1 / L0)
Substituting the values, we get:
T1 / T0 = √(13 * L0 / L0) = √13
Since the pendulum keeps perfect time, the ratio of the periods is equal to the ratio of the time intervals. Therefore, the new time on the clock after 24 hours will be:
New Time = 24 hours * (√13)
Performing the calculation, we get:
New Time = 24 * √13 = 24 * 3.60555 = 86.5333 hours
Rounding to five-digit precision, the time on the clock will read approximately 86.5333 hours after the pendulum length is increased.
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A sphere of radius 2.00cm with a black surface is cooled and then suspended in a large evacuated enclosure with black walls maintained at 27°C. If the rate of change of thermal energy of sphere is 1.85J/s when its temperature is -73°C, calculate the value of Stefan's constant.
To calculate the value of Stefan's constant, we can utilize the formula for the rate of change of thermal energy. The rate of change of thermal energy is given by dQ/dt = εσA(T^4 - T_env^4), where dQ/dt is the rate of change of thermal energy, ε is the emissivity of the surface, σ is Stefan's constant.
A is the surface area of the sphere, T is the temperature of the sphere in Kelvin, and T_env is the temperature of the environment in Kelvin. Given that the rate of change of thermal energy is 1.85 J/s when the temperature of the sphere is -73°C, we first need to convert the temperature to Kelvin. -73°C is equivalent to 200.15 K. Substituting the values into the formula, we have 1.85 = εσA((200.15)^4 -(27 + 273.15)^4). Rearranging the equation, we get σ = (1.85) / (εA((200.15)^4 - (27 + 273.15)^4)) To calculate the value of Stefan's constant, we need to know the emissivity and surface area of the sphere. Please provide the emissivity and surface area of the sphere so that we can continue the calculation.
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A circuit is constructed with five resistors and a battery as shown. The values for the resistors are: R1 = R5 = 66 2, R2 = 136, R3 = 95 02, and R4 = 112 2. The battery voltage is V = 12 V. ww R5 1) What is Rab, the equivalent resistance between points a and b? Submit 2) What is Rac, the equivalent resistance between points a and c? Submit 3) What is 15, the current that flows through resistor R5? A Submit 4) What is 12, the current that flows through resistor R2? Submit 5) What is 11, the current that flows through the resistor R₁? A Submit 6) What is V4, the voltage across resistor R4? V Submit ď + + (+) (+) +
1) The Rab is 34.785 Ω, when the equivalent resistance between points a and b.
2) The Rac is 409.42 Ω , when the equivalent resistance between points a and c.
3) The I5 is 0.1815 A, when the current that flows through resistor R5.
4) The I2 is 0.0293 A, when the current that flows through resistor R2.
5) The I1 is 0.0293 A, when the current that flows through the resistor R₁.
6) The V4 = 3.2634 V, when the voltage across resistor R4.
To solve the circuit and answer the given questions, we need to apply the principles of series and parallel resistors.
To calculate the equivalent resistance between points a and b (Rab), we need to consider R1, R2, and R3 which are in parallel. The reciprocal of the equivalent resistance is given by the sum of the reciprocals of the individual resistances:
1/Rab = 1/R1 + 1/R2 + 1/R3
1/Rab = 1/66.2 + 1/136 + 1/95.02
Calculating the sum on the right side and taking the reciprocal, we get:
Rab = 34.785 Ω
To find the equivalent resistance between points a and c (Rac), we consider R1, R2, R3, and R4 which are in series. The sum of their resistances gives us Rac:
Rac = R1 + R2 + R3 + R4
Rac = 66.2 + 136 + 95.02 + 112.2
Rac = 409.42 Ω
The current flowing through resistor R5 can be determined using Ohm's Law (I = V/R):
I5 = V / R5
I5 = 12 / 66.2
I5 = 0.1815 A
The current flowing through resistor R2 is equal to the total current in the circuit, which can be calculated using the equivalent resistance Rac and Ohm's Law:
I2 = V / Rac
I2 = 12 / 409.42
I2 = 0.0293 A
The current flowing through resistor R1 is also equal to the total current in the circuit:
I1 = V / Rac
I1 = 12 / 409.42
I1 = 0.0293 A
The voltage across resistor R4 (V4) can be calculated using Ohm's Law:
V4 = I2 * R4
V4 = 0.0293 * 112.2
V4 = 3.2634 V
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A wheel starts from rest and accelerates with constant angular acceleration to an angular velocity of 900 rev.min' in 20 s. At the end of 1s a) find the angle through which the wheel has rotated;
and b) compute and show in a diagram the magnitude and
direction of the tangential and radial components of acceleration of a point 0.2 m from the axis.
The angle through which the wheel has rotated at the end of 1 second is approximately 30 degrees.
To find the angle through which the wheel has rotated, we can use the formula:
θ = ωi * t + (1/2) * α * t^2,
where θ is the angle, ωi is the initial angular velocity, t is the time, and α is the angular acceleration.
Given that the initial angular velocity is 0 (starting from rest) and the angular acceleration can be calculated using the final angular velocity and the time:
α = (ωf - ωi) / t,
where ωf is the final angular velocity and t is the time.
In this case, the final angular velocity is 900 rev/min, which can be converted to radians per second:
ωf = (900 rev/min) * (2π rad/rev) * (1 min/60 s) ≈ 94.25 rad/s.
Substituting the values into the equation, we have:
θ = 0 * 1 + (1/2) * α * (1)^2.
Evaluating this expression, we find:
θ ≈ 0.5 * α.
Therefore, the angle through which the wheel has rotated at the end of 1 second is approximately 0.5 times the angular acceleration.
) To compute the magnitude and direction of the tangential and radial components of acceleration of a point 0.2 m from the axis, we can use the following formulas:
Tangential acceleration (at) = r * α,
Radial acceleration (ar) = r * ω^2,
where r is the radial distance from the axis, α is the angular acceleration, and ω is the angular velocity.
Given that r = 0.2 m, we can substitute the values and calculate the magnitudes:
Tangential acceleration (at) = (0.2 m) * α,
Radial acceleration (ar) = (0.2 m) * (ω^2).
To determine the direction of these components, we refer to the diagram. The tangential acceleration is in the direction of the tangential velocity (tangent to the circular path), and the radial acceleration is perpendicular to the tangential acceleration and directed toward the center of the circular path.
Therefore, in the diagram, draw an arrow representing the tangential acceleration in the direction of the tangential velocity and another arrow representing the radial acceleration pointing toward the center of the circle.
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Regenerate response
Answer question related to circuit given [10 pts] Given vs(t) = 15 cos(100t) V US(1) a) Write vs(t) in phasor form b) In Figure 4a, what is Z₁? Figure 4a Z₁ c) In Figure 4a, what is Ze? 2₁ Figure 4b Note: Figure 4b is equivalent of Figure 4a as follows: d) In Figure 4b, Z₁ = 100, let Z₂ = Z₁ (found in part (b)), and let Z3 (1502 resistor in parallel with Ze (found in part (c))). Find Z3 in polar form. Show work, box answer. ) Compute Zeq = Z₁ + Z₂ + Z3 in polar form. f) Compute current I in Figure 4b using V as value obtained in part (a) and Zeq obtained in part (e). Show all work, final answer should be in phasor form. Write units and box answer. 6/6 10 f www m 50 mH 150 (1) 12(1) 1mF 2₂
The phasor representation of the voltage source vs(t) = 15 cos(100t) V is 15∠0° V.
What is the phasor representation of the voltage source vs(t) = 15 cos(100t) V in the given circuit?a) What is the phasor form of the voltage source vs(t) = 15 cos(100t) V?
b) In Figure 4a, what is the value of Z₁?
c) In Figure 4a, what is the equivalent impedance Ze?
d) In Figure 4b, if Z₁ = 100 and Z₂ is the same as Z₁, and Z₃ is a 150 Ω resistor in parallel with Ze, what is the value of Z₃ in polar form?
e) Compute the total equivalent impedance Zeq = Z₁ + Z₂ + Z₃ in polar form.
f) Compute the current I in Figure 4b using the voltage V obtained in part (a) and the equivalent impedance Zeq obtained in part (e). Provide the final answer in phasor form.
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Your frind is driving at 20 miles per hour, then moving to another road, you increase the speed to 60 miles per hour. At 20 mph the car can be stopped in 10 meters. How far in meters does the car move before coming to a stop at the new speed? Round to the nearest hundredth.)
To solve this problem, we need to first convert the speeds from miles per hour to meters per second, as well as convert the distance from miles to meters.
1 mile = 1609.34 meters
1 hour = 3600 seconds
Converting the speeds:
20 miles per hour = (20 * 1609.34) / 3600 meters per second ≈ 8.94 m/s
60 miles per hour = (60 * 1609.34) / 3600 meters per second ≈ 26.82 m/s
Now, let's calculate the distance traveled before coming to a stop at the new speed.
At 20 mph, the car can be stopped in 10 meters. So, the deceleration rate can be calculated as follows:
Initial velocity (u) = 8.94 m/s
Final velocity (v) = 0 m/s
Distance (s) = 10 meters
Using the equation of motion: v^2 = u^2 + 2as
0 = (8.94)^2 + 2a(10)
Solving for acceleration (a):
a = - (8.94)^2 / (2 * 10) ≈ -3.978 m/s^2 (negative sign indicates deceleration)
Now, let's calculate the distance traveled at 60 mph using the new speed and the calculated deceleration rate.
Initial velocity (u) = 26.82 m/s
Final velocity (v) = 0 m/s
Acceleration (a) = -3.978 m/s^2
Using the equation of motion: v^2 = u^2 + 2as
0 = (26.82)^2 + 2(-3.978)s
Solving for distance (s):
s ≈ [(26.82)^2] / (2 * 3.978) ≈ 181.03 meters
Therefore, the car moves approximately 181.03 meters before coming to a stop at the new speed.
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A light ray traveling from the Sun reaches the eye of a fish, see the image. What is the angle of refraction if the angle of incidence is 70°? Mark with an "x " the apparent position of the Sun according to the fish and explain why the fish believes that the Sun is located at that spot. 70° air 70° water
The angle of refraction is approximately 48.2°. The angle of refraction can be determined using Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the velocities of light in the two media. Mathematically, it can be expressed as:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
where n₁ and n₂ are the refractive indices of the respective media, θ₁ is the angle of incidence, and θ₂ is the angle of refraction.
In this case, the light ray is traveling from air to water, and the angle of incidence is given as 70°. The refractive indices of air and water are approximately 1 and 1.33, respectively.
Using Snell's law, we can calculate the angle of refraction as follows:
1 * sin(70°) = 1.33 * sin(θ₂)
Solving for θ₂, we find:
θ₂ ≈ 48.2°
Therefore, the angle of refraction is approximately 48.2°.
As for the apparent position of the Sun according to the fish, it will be seen as the direction from which the refracted light rays appear to originate. In this case, the apparent position of the Sun would be marked with an "x" at a location where the refracted rays converge after passing through the water. This apparent position is different from the actual position of the Sun in the sky.
The fish perceives the apparent position of the Sun at that spot because the fish's eye receives the refracted light rays from different directions and interprets them as if they were coming from a single source. Due to the bending of light at the air-water interface, the fish's eye receives light rays from different angles, and the apparent position of the Sun is determined by the direction from which the light rays appear to converge. This creates an optical illusion for the fish, making it believe that the Sun is located at the apparent position marked by the "x".
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Part B If a block is moving to the left at a constant velocity, what can one conclude? ► View Available Hint(s) O There is exactly one force applied to the block. O The net force applied to the block is directed to the left. O The net force applied to the block is zero. O There must be no forces at all applied to the block. Submit Part C A block of mass 2 kg is acted upon by two forces: 3 N (directed to the left) and 4 N (directed to the right). What can you say about the block's motion? View Available Hint(s) O It must be moving to the left. O It must be moving to the right O It must be at rest O It could be moving to the left, moving to the right, or be instantaneously at rest Submit Part D A massive block is being pulled along a horizontal frictionless surface by a constant horizontal force. The block must be View Available Hint(s) O continuously changing direction O moving at constant velocity moving with a constant nonzero acceleration A O moving with continuously increasing acceleration Submit Part E Two forces, of magnitude 4 N and 10 N, are applied to an object. The relative direction of the forces is unknown. The net force acting on the object Check all that apply. View Available Hint(s) cannot have a magnitude equal to 5 N cannot have a magnitude equal to 10 N cannot have the same direction as the force with magnitude 10 N must have a magnitude greater than 10 N Submit
For a block moving to the left at a constant velocity, it can be concluded that the net force applied to the block is zero.
When a block is moving to the left at a constant velocity, it means that the forces acting on the block are balanced. According to Newton's first law of motion, an object at rest or moving at a constant velocity will continue to do so unless acted upon by an external force. In this case, since the block is moving at a constant velocity to the left, it means that the net force applied to the block is zero.
In the second scenario, where a block of mass 2 kg is acted upon by two forces, 3 N (directed to the left) and 4 N (directed to the right), the net force can be calculated by subtracting the force acting in the opposite direction. In this case, the net force would be 3 N - 4 N = -1 N. Since the net force is directed to the left, the block's motion would be towards the left.
In the third scenario, where a massive block is being pulled along a horizontal frictionless surface by a constant horizontal force, the block would be moving at a constant velocity. This is because there is no friction acting on the block to oppose its motion, and the constant horizontal force provides the necessary balanced force to maintain a constant velocity.
In the fourth scenario, the magnitude and relative direction of the forces applied to an object determine the net force. The net force is the vector sum of the individual forces. In this case, the net force cannot have a magnitude equal to 5 N or 10 N since the forces are not in the same direction. The net force must have a magnitude greater than 10 N, as it is the sum of two forces. The direction of the net force depends on the relative direction of the two forces applied.
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A 10 cm length of wire carries a current of 2.0 A in the positive x-direction. The force on this wire due to the presence of a magnetic field B
is F
=(3.0
^
+2.0 k
^
)N. If this wire is now rotated so that the current is in the positive y-direction, the force on the wire is F
=(−2.0
^
−3.0 k
^
) N. Determine the magnetic field B
in unit vector form. [6]
the magnetic field B can be expressed as B = (-11-- 0- 0) N.To determine the magnetic field B in unit vector form, we can use the given forces and the right-hand rule for the magnetic force on a current-carrying wire.
First, let's consider the force on the wire when the current is in the positive x-direction. The given force is F = (3.0+ 2.0) N. The force experienced by a wire in a magnetic field is given by the equation F = × , where is the current, is the length of the wire, and is the magnetic field. Since the force is in the positive x-direction, and the current is also in the positive x-direction, the magnetic field B should be in the negative y-direction.
Next, let's consider the force on the wire when the current is in the positive y-direction. The given force is F = (−2.0− 3.0) N. Since the force is in the negative y-direction, and the current is now in the positive y-direction, the magnetic field B should be in the negative x-direction.
Therefore, the magnetic field B can be expressed as B = (-1- 0- 0) N.
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What is the maximum electric field strength in an electromagnetic wave that has a maximum magnetic 1 field strength of 5.5 x 10-4T? Use c for the speed of light. νεομο
The maximum electric field strength in the electromagnetic wave is 1.65 x 10^5 V/m.
In an electromagnetic wave, the electric field strength (E) and the magnetic field strength (B) are related by the equation:
E = c * B
where c is the speed of light in vacuum, and B is the maximum magnetic field strength.
Given that the maximum magnetic field strength (B) is 5.5 x 10^-4 T, we can calculate the maximum electric field strength (E) using the speed of light (c).
The speed of light in vacuum is approximately 3.00 x 10^8 m/s.
E = c * B
= (3.00 x 10^8 m/s) * (5.5 x 10^-4 T)
= 1.65 x 10^5 V/m
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You are riding a train traveling North at 7.2 m/s relative to the ground. You decide you want a snack, so you walk towards the food car. If your speed relative to the ground was 5.9 m/s, how fast were you walking relative to the train and in which direction? O 13.1 m/s South O 1.3 m/s South O 13.1 m/s North O 1.3 m/s North
You were walking at a speed of 1.3 m/s North relative to the train. When you walk towards the food car on a moving train, your speed relative to the ground is the vector sum of your speed relative to the train and the train's speed relative to the ground.
In this case, the train is traveling North at 7.2 m/s, and you are walking with a speed of 5.9 m/s relative to the ground. To find your speed relative to the train, you need to subtract the train's velocity from your velocity relative to the ground. Since you were walking in the same direction as the train, your speed relative to the train is 5.9 m/s minus 7.2 m/s, which gives a result of -1.3 m/s. The negative sign indicates that you were walking in the opposite direction of the train's motion. Therefore, you were walking at a speed of 1.3 m/s relative to the train, in the opposite direction of the train's motion, which is North.
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12.0, the vector B with components B, = 4.00 and By = -4.00, and D= 0- Given the vector A with components A, = 4.00 and A, the vector D=A- B, calculate the magnitude D of the vector D. Determine the angle 0 that the vector D makes with respect to the positive x-axis.
The magnitude of vector D is approximately 8.94.
The angle θ that vector D makes with respect to the positive x-axis is approximately -26.57°.
To calculate the vector D = A - B, we subtract the corresponding components of A and B:
Dx = Ax - Bx = 4.00 - 12.0 = -8.00
Dy = Ay - By = 0 - (-4.00) = 4.00
The magnitude of vector D is calculated using the Pythagorean theorem:
|D| = sqrt(Dx^2 + Dy^2) = sqrt((-8.00)^2 + 4.00^2) = sqrt(64.00 + 16.00) = sqrt(80.00) ≈ 8.94
To calculate the angle θ that vector D makes with respect to the positive x-axis, we can use the inverse tangent function:
θ = arctan(Dy / Dx) = arctan(4.00 / -8.00) ≈ -26.57°
Note: The angle is negative because the vector D lies in the fourth quadrant with a negative x-component.
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A converging lens has a focal length of 20.0 cm. Locate the images for each of the following object distances. (Enter 'infinity' for the image distance if necessary.) For each case, state whether the image is real or virtual and upright or inverted. Find the magnification. (If there is no answer for a blank enter N/A.) (a) 40.0 cm 2 X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. cm in front of the lens Oreal, inverted O virtual, inverted O no image formed O real, upright O virtual, upright magnification 2 X The response you submitted has the wrong sign.x (b) 20.0 cm cm-location of the image-- cm-location of the image-- no image formed O real, inverted O real, upright O virtual, inverted O virtual, upright magnification (c) 10.0 cm O inverted, real O inverted, virtual O erect, virtual erect, real O no image formed magnification X
(a) For an object distance of 40.0 cm in front of the converging lens, the image is real, inverted, and the magnification is 2X. (b) For an object distance of 20.0 cm in front of the converging lens, no image is formed.
(c) For an object distance of 10.0 cm in front of the converging lens, the image is erect and virtual.
(a) For an object distance of 40.0 cm:
Using the lens formula, 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance.
Substituting the given values, we have 1/20 cm = 1/40 cm + 1/di.
Solving for di, we find di = 13.3 cm.
Since di is positive, the image is real. Since the magnification, M = -di/do = -13.3 cm / 40.0 cm = -0.333, the image is inverted.
(b) For an object distance of 20.0 cm:
Using the lens formula, we find 1/20 cm = 1/20 cm + 1/di.
Simplifying, we get 0 = 1/di.
Since di is infinite, no image is formed.
(c) For an object distance of 10.0 cm:
Using the lens formula, we find 1/20 cm = 1/10 cm + 1/di.
Simplifying, we get 1/di = 1/20 cm - 1/10 cm = -1/20 cm.
Since 1/di is negative, the image is virtual. The magnification, M = -di/do = -(-1/20 cm) / 10.0 cm = 0.05, so the image is erect.
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Will begins his spin with his leg and arms extended. Pulling his leg and arms in, his rotational inertia is reduced by 1/3. By what factor is his angular velocity changed?
The angular velocity is changed by a factor of 3/2 (or 1.5).According to the law of conservation of angular momentum, when there is no external torque acting on a system, the total angular momentum remains constant. In this case, as Will pulls his leg and arms in, reducing his rotational inertia by 1/3, his angular momentum must remain constant.
Since angular momentum (L) is given by the product of rotational inertia (I) and angular velocity (ω), we can write L = Iω.
If the rotational inertia is reduced by 1/3, it means the new rotational inertia (I') is 2/3 of the original value. Therefore, we have (2/3)Iω' = Iω, where ω' is the new angular velocity.
Simplifying the equation, we find ω' = (3/2)ω.
Hence, thethe angular velocity is changed by a factor of 3/2 (or 1.5).
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An ambulance moving at 30.0 m/s is sounding its siren with a frequency of 512 Hz. If the speed of a sound in air is 344 m/s, what is the wavelength of sound "behind" the ambulance?
0.79 m
0.61 m
0.73 m
0.67 m
The wavelength of sound "behind" the ambulance is approximately 0.672 m.
To calculate the wavelength of sound "behind" the ambulance, we can use the formula:
wavelength = speed of sound / frequency
Given:
Speed of sound in air (v) = 344 m/s
Frequency of the siren (f) = 512 Hz
Plugging in these values into the formula, we have:
wavelength = 344 m/s / 512 Hz
Calculating the result:
wavelength ≈ 0.672 m
Therefore, the distance between neighbouring identical portions of a sound wave is its wavelength. The distance between adjacent compressions in a sound wave is the same as the distance between adjacent crests in a transverse wave—both are separated by one wavelength.
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