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V125BC [204]
3 years ago
14

Which phrase best describes wave speed?

Physics
2 answers:
VladimirAG [237]3 years ago
8 0

Explanation:

In the case of a wave, the speed is the distance traveled by a given point on the wave (such as a crest) in a given interval of time. In equation form, If the crest of an ocean wave moves a distance of 20 meters in 10 seconds, then the speed of the ocean wave is 2.0 m/s.

Ksju [112]3 years ago
6 0
I think the answer is c
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They both flow in currents. Water has a pump that works like a battery and pipes that work like a circuit.

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If the sun is 400 times bigger than the moon, how couild the moon possibly cover the sun during a solar eclipse?​
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Explanation:

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IP Force to Hold a Baseball A person holds a 1.42-N baseball in his hand, a distance of 34.0 cm from the elbow joint, as shown i
mixer [17]

We will assume that the CM of the arm is at "L" from the elbow, and the ball is at 34cm. Then the net torque is computed by:


Net τ = 1.42 N * 34 cm + 1.50 kg * 9.8m/s² * 34 cm/2 – 12.6 N*2.75cm 

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A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t
Irina-Kira [14]

Answer:

part (a) \alpha\ =\ -2.5\ rad/s^2

part (b) N = 79.61 rev

part (c) \tau\ =\ 23.54\ Nm

Explanation:

Given,

  • Initial speed of the wheel = w_o\ =\ 50.0\ rad/s
  • total time taken = t = 20.0 sec

part (a)

Let \alpha be the angular acceleration of the wheel.

Wheel is finally at the rest. Hence the final angular speed of the wheel is 0.

\therefore w_f\ =\ w_0\ +\ \alpha t\\\Rightarrow \alpha\ =\ -\dfrac{w_0}{t}\\\Rightarrow \alpha\ =\ -\dfrac{50}{20}\\\Rightarrow \alpha\ =\ -2.5\ rad/s^2

part (b)

Let \theta be the total angular displacement of the wheel from initial position till the rest.

\therefore \theta\ =\ w_0t\ +\ \dfrac{1}{2}\alphat^2\\\Rightarrow \theta\ =\ 50\times 20\ -\ 0.5\times 2.5\times 20^2\\\Rightarrow \theta\ =\ 500\ rad

We know,  1 revolution = 2\pi rad

Let N be the number of revolution covered by the wheel.

\therefore N\ =\ \dfrac{\theta}{2\pi}\\\Rightarrow N\ =\ \dfrac{500}{2\times 3.14}\\\Rightarrow N\ =\ 79.61\ rev

Hence the 79.61 revolution is covered by the wheel in the 20 sec.

part (c)

Given,

  • Mass of the pole = m = 4 kg
  • Length of the pole = L = 2.5 m
  • Angle of the pole with the horizontal axis = \theta\ =\ 60^o

Now the center of mass of the pole = d\ =\ \dfra{L}{2}\ =\ \dfrac{2.5}{2}\ =\ 1.25\ m

Weight component of the pole perpendicular to the center of mass = F\ =\ mgcos\theta

\therefore \tau\ =\ F\times d\\\Rightarrow \tau\ =\ 4\times 9.81\times cos60^o\times 1.25\\\Rightarrow \tau\ =\ 23.54\ Nm

3 0
3 years ago
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