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3 years ago

How are motors and generators different?

1 answer:

4 0

A generator converts mechanical energy into electrical energy, while a motor does the opposite - it converts electrical energy into mechanical energy

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(a) Write an equation describing a sinusoidal transverse wave traveling on a cord in the positive of a y axis with an angular wa

Vedmedyk [2.9K]

Missing data: the wave number

$k=60 cm^{-1}$

(a) $z = 0.003 sin (6000y-31.4 t)$

For a transverse wave travelling in the positive y-direction and with vibration along the z-direction, the equation of the wave is

$z = A sin (ky-\omega t)$

where

A is the amplitude of the wave

k is the wave number

$\omega$ is the angular frequency

t is the time

In this situation:

A = 3.0 mm = 0.003 m is the amplitude

$k = 60 cm^{-1} = 6000 m^{-1}$ is the wave number

$T = 0.20 s$ is the period, so the angular frequency is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.20}=31.4 rad/s$

So, the wave equation (in meters) is

$z = 0.003 sin (6000y-31.4 t)$

(b) 0.094 m/s

For a transverse wave, the transverse speed is equal to the derivative of the displacement of the wave, so in this case:

$v_t = z' = -A \omega cos (ky-\omega t)$

So the maximum transverse wave occurs when the cosine term is equal to 1, therefore the maximum transverse speed must be

$v_{t}_{max} =\omega A$

where

$\omega = 31.4 rad/s\\A = 0.003 m$

Substituting,

$v_{t}_{max}=(31.4)(0.003)=0.094 m/s$

(c) 5.24 mm/s

The wave speed is given by

$v=f \lambda$

where

f is the frequency of the wave

$\lambda$ is the wavelength

The frequency can be found from the angular frequency:

$f=\frac{\omega}{2\pi}=\frac{31.4}{2\pi}=5 Hz$

While the wavelength can be found from the wave number:

$\lambda = \frac{2\pi}{k}=\frac{2\pi}{6000}=1.05\cdot 10^{-3} m$

Therefore, the wave speed is

$v=(5)(1.05\cdot 10^{-3} )=5.24 \cdot 10^{-3} m/s = 5.24 mm/s$

7 0

4 years ago

If you were to travel from the equator to the higher latitudes (near either pole) on Saturn, the differential rotation would cau

Dmitrij [34]

Answer:

Saturn's differential rotation will cause the length of a day measures to be longer by 0.4 hours

Explanation:

Differential rotation occurs due to the difference in angular velocities of an object as we move along the latitude of the or as we move into different depth of the object, indicating the observed object is in a fluid form

Saturn made almost completely of gas and has differential motion given as follows

Rotation at the equator = 10 hours 14 minutes

Rotation at high altitude = 10 hours 38 minutes

Therefore;

The differential rotation = 10 hours 38 minutes - 10 hours 14 minutes

The differential rotation = 24 minutes = 24 minutes × 1 hour/(60 minutes) = 0.4 hours

The differential rotation = 0.4 hours

Therefore, the measured day at the higher altitude will be 0.4 longer than at the equator.

8 0

3 years ago

Can anybody write a short poem about friction

Luden [163]

Answer:

you will be the clouds
and I will be the sky.
you will be the ocean
and I will be the shore.
you will be the trees
and I will be the wind.

whatever we are, you and I will always collide.

There you go! Let me know if it helped.
:)

8 0

2 years ago

The weight of a luggage is 69.3 N on the moon. Find its weight on the Earth.

Galina-37 [17]

Answer:

Explanation:

weight on moon = 1/6* weight on earth

69.3=1/6*weight on earth

weight on earth = 69.3*6

weight on earth = 415.8 N

8 0

3 years ago

The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1

Vaselesa [24]

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

$F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2})$ (1)

Where:

$F$ - Explosive force, measured in newtons.

$\Delta s$ - Barrel length, measured in meters.

$m$ - Mass of the shell, measured in kilograms.

$v_{o}$ , $v_{f}$ - Initial and final speeds of the shell, measured in meters per second.

If we know that $m = 1250\,kg$ , $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 750\,\frac{m}{s}$ and $\Delta s = 15\,m$ , then the explosive force experienced by the shell inside the barrel is:

$F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}$

$F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}$

$F = 23437500\,N$

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

6 0

3 years ago