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Scrat [10]
3 years ago
5

Through which of these media do sound waves travel most slowly? A.iron. B.Wood. C.Water. D.Cold Air

Physics
1 answer:
EleoNora [17]3 years ago
5 0
The answer should be D) Cold air because even though its true sound can travel through all types of matter, air which is a gas, can travel but it travels SLOWLY while sound travels quickly in SOLIDS.
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20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What is the new vol
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Initial volume of the gas (V1) = 10 inches^3
Initial pressure (P1) = 5 psi
Final pressure after compression of the gas (P1) = 10 psi
Let us assume the final volume of the gas (V2) = x
According to Boyle's Gas law, the pressure and volume of a gas remains constant under ideal condition. Then
P1V1= P2V2
5 * 10 = 10 * x
50 = 10x
x = 50/10
   = 5 cubic inches
So the volume of the gas after it was compressed was 5 cubic inches. I hope the procedure is clear enough for you to understand.
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3 years ago
A builder drops a brick from a height of 15 m above the ground. The gravitational field strength g is 10 N/ kg. What is the spee
Basile [38]

The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

  • Initial Velocity; u = 0
  • Height from which it has dropped; h = 15m
  • Gravitational field strength; g = 10N/kg = 10 \frac{kg.m/s^2}{kg} = 10m/s^2

Final speed of brick as it hits the ground; v =  \ ?

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

v^2 = u^2 + 2gh

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Learn more about equations of motion: brainly.com/question/18486505

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