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GREYUIT [131]
4 years ago
7

A constant net torque is exerted on an object. Which of the following quantities for the object cannot be constant? (Select all

that apply.)
A. kinetic energy
B. angular velocity
C. angular acceleration
D. moment of inertia
E. angular position
Physics
1 answer:
prohojiy [21]4 years ago
3 0

Answer:

A. kinetic energy

B. angular velocity

E. angular position

Explanation:

The quantities that cannot be constant if a constant net torque is exerted on an objecta are:

A. Kinetic energy. If a torque is applied, the linear or angular speed will be changing at a rate proportional to the torque, so the kinetic energy will change too.

B. Angular velocity. It will change at a rate equal to the torque.

C. Angular position. If the angular velocity changes, the angular position will change.

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Given that y is the vertical displacement of a wave at anytime and A is the amplitude of the wave, w is the angular speed of the
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Explanation:

The general equation describing a wave is:

y(x,t) = A sin(kx - wt)

Let's say that for a particular wave on a string the equation is:

y(x,t) = (0.9 cm) sin[(1.2 m-1)x - (5.0 s-1)t]

(a) Determine the wave's amplitude, wavelength, and frequency.

(b) Determine the speed of the wave.

(c) If the string has a mass/unit length of m = 0.012 kg/m, determine the tension in the string.

(d) Determine the direction of propagation of the wave.

(e) Determine the maximum transverse speed of the string.

Solutions

Part (a): The wave's amplitude, wavelength, and frequency can be determined from the equation of the wave:

y(x,t) = (0.9 cm) sin[(1.2 m-1)x - (5.0 s-1)t]

The amplitude is whatever is multiplying the sine.

A = 0.9 cm

The wavenumber k is whatever is multiplying the x:

k = 1.2 m-1The wavelength isl=2pk= 5.2 m

The angular frequency w is whatever is multiplying the t.

w = 5.0 rad/sf=w2p= 0.80 Hz

Part (b): The wave speed can be found from the frequency and wavelength:

v = f l = 0.80 * 5.2 = 4.17 m/s

Part (c): With m = 0.012 kg/m and the wave speed given by:v=(Tm)½

This gives a tension of T = m v2 = 0.012 (4.17)2 = 0.21 N.

Part (d): To find the direction of propogation of the wave, just look at the sign between the x and t terms in the equation. In our case we have a minus sign:

y(x,t) = (0.9 cm) sin[(1.2 m-1)x - (5.0 s-1)t]

A negative sign means the wave is traveling in the +x direction.

A positive sign means the wave is traveling in the -x direction.

Part (e): To determine the maximum transverse speed of the string, remember that all parts of the string are experiencing simple harmonic motion. We showed that in SHM the maximum speed is:

vmax = Aw

In this case we have A = 0.9 cm and w = 5.0 rad/s, so:

vmax = 0.9 * 5.0 = 4.5 cm/s

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