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S_A_V [24]
3 years ago
10

A surface wave is a combination of what two waves?

Physics
2 answers:
Elena-2011 [213]3 years ago
8 0

A for sure, just took the quiz:)

kotegsom [21]3 years ago
4 0

transverse and longitudinal

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An object is dropped at a height of 12 m from the ground. How fast is it moving just before it hits the ground?
sp2606 [1]

The final speed of the object is 15.3 m/s

Explanation:

The object in the problem is moving by uniformly accelerated motion (it is in free fall), so we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a=g=9.8 m/s^2 is the acceleration of gravity (taking downward as positive direction)

s is the vertical displacement

For the object in this problem:

u = 0 (it starts from rest)

s = 12 m is the displacement of the object

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(12)}=15.3 m/s

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3 years ago
A box slides down a frictionless incline, gaining speed. The work done by the normal force n is _______.
jeka57 [31]

The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

<h3>What is normal force?</h3>

The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.

This force is applied by the solid bodies on each other in order to prevent the passing through each other.

A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.

  • The body is gaining the speed, which means there is a change in kinetic energy.
  • The change in kinetic energy is equal to the work done.
  • The friction force is the product of coefficient of the friction and normal force.
  • The friction force for the given case is zero. Thus, the normal force must be equal to the zero.

Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

Learn more about the normal force here;

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