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Savatey [412]
3 years ago
10

As ocean waves get closer to the shore they slow down. This slowing causes them to bend and come in close to parallel with the s

hore. This is because shallow water and deep water can be thought of as two different media. This bending of a wave is the wave interaction known as
Physics
2 answers:
olchik [2.2K]3 years ago
6 0
The answer is Refraction which is the bending 
Inga [223]3 years ago
3 0
Refraction is the right answer.
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If the treadmill’s initial speed is 1.7m/s, what will its final speed be?
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A 0.26 kg rod of length 80 cm is suspended by a frictionless pivot at one end. It is held horizontal and released.
Daniel [21]

Answer:

a) a_{center} = 7.38 ~m/s^2

b) a_{end} = 14.77 ~m/s^2

c) v_{center} = 2.43~m/s

Explanation:

a) Immediately after the rod is released, <u>the rod is still horizontal but now subject to gravity.</u> Since one end of the rod is fixed, then the weight of the rod applies a torque. Then by Newton's Second Law, the acceleration can be found.

\tau =I\alpha

where I is the moment of inertia of the rod with respect to its fixed end, and α is the angular acceleration.

The net torque of the rod is

\vec{\tau} = \vec{r} \times \vec{F}\\\tau = rF\sin(90) = rF

where r is the distance from center of the mass to the fixed end, so r = 0.4 m.

The weight of the rod is w = mg = 0.26 x 9.8 = 2.54 N.

So the net torque is τ = 1.01 Nm.

The moment of inertia of the rod is

I = \frac{1}{3}mL^2 = \frac{1}{3}(0.26)(0.8)^2 = 0.055~kg m^2

So, the Newton's Second Law yields

\tau = I\alpha\\\alpha = \frac{\tau}{I} = \frac{1.01}{0.055} = 18.47

<u>The relation between angular acceleration and linear acceleration is a = αr </u>

So, the linear acceleration of the rod is

a = \alpha r = 7.38~m/s^2

b) Using the same relationship between angular acceleration and linear acceleration, the linear acceleration of the end of the rod can be found.

a = \alpha L = 14.77~m/s^2

c) The conservation of energy can be used to find the velocity when the rod is vertical.

K_1 + U_1 = K_2 + U_2\\0 + mg(L/2) = \frac{1}{2}I\omega^2 + 0\\(0.26)(9.8)(0.4) = \frac{1}{2}(0.055)\omega^2\\\omega = 6.08~rad/s

The linear velocity is v = ωr, so

v = 2.43 m/s.

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