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Gwar [14]
3 years ago
9

What do ocean waves and sound waves have in common?

Physics
2 answers:
Anettt [7]3 years ago
7 0
Since they are both examples of moving waves, they both transmit energy.
eduard3 years ago
6 0
They both make sound.
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a surface recieving sound is moved from it original position to a position three times farther away from the source of the sound
Delicious77 [7]
Sound intensity = 1/(r^2)

That is Sound intensity is indirectly proportional to  the distance. Therefore, sound becomes 9 times less intense.
7 0
2 years ago
A 438kg car is accelerating east at 2.55m/s^2. What is the total force acting east on the car
lisabon 2012 [21]

Answer:

<h2>1116.9 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 438 × 2.55

We have the final answer as

<h3>1116.9 N</h3>

Hope this helps you

5 0
3 years ago
a teacher pushed a 10kg desk across a floor for a distance of 5m. she exerted a horizontal force of 20n. how much work was done?
serious [3.7K]
Work Done = Force x distance
Since she exerted a horizontal force of 20N over a distance of 5m, the work done is 20N x 5m which is equals to 100 joules
7 0
3 years ago
A 0.14-MIN baseball is dropped from rest. It has a momentum of 0.90 kg⋅m/skg⋅m/s just before it lands on the ground.
nikitadnepr [17]

The time spent in the air by the ball at the given momentum is 6.43 s.

The given parameters;

  • <em>momentum of the ball, P = 0.9 kgm/s</em>
  • <em>weight of the ball, W = 0.14 N</em>

The impulse experienced by the ball is calculated as follows;

Ft = \Delta P

where;

Ft is impulse

\Delta P is change in momentum

The time of motion of the ball is calculated as follows;

t = \frac{\Delta P}{F} \\\\t = \frac{0.9 - 0}{0.14} \\\\t = 6.43 \ s

Thus, the time spent in the air by the ball at the given momentum is 6.43 s.

Learn more here:brainly.com/question/13468390

7 0
2 years ago
Who can help me?? physic question​
IgorC [24]

Assuming acceleration due to gravity of the moon is constant and there’s no initial velocity in the mans jump we can use one of the kinematic equations. x(final)=x(initial)+(1/2)gt^2. Plug in known values. 0=10-(1.62/2)t^2. The value 1.62 is acceleration of gravity on the moon. Now simply solve for t. t=3.513

5 0
2 years ago
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