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matrenka [14]
4 years ago
6

What is The common value for the speed of light in a vacuum

Physics
1 answer:
Talja [164]4 years ago
4 0
The most common value for the speed of light is 3*10^8 meters/second.

A more accurate number is <span>299 792 458 m/second, but that number is hardly ever used.</span>
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An Echo is an example of which type of wave interaction
maksim [4K]

An echo is an example of the <em>reflection</em> of sound waves.

7 0
3 years ago
A 40kg rock is perched atop a cliff that is 50 m high. What is it’s potential energy?
LekaFEV [45]
Potential energy, is energy due to its position.

Given: Mass m = 40 Kg;  Height h = 50 m

Required: Potential energy P.E = ?

Formula:  P.E = mgh   P.E = (40 Kg)(9.8 m/s²)(50 m)

                P.E = 19,600 J
4 0
4 years ago
15) What is the frequency of a pendulum that is moving at 30 m/s with a wavelength of .35 m?
____ [38]

A pendulum is not a wave.

-- A pendulum doesn't have a 'wavelength'.

-- There's no way to define how many of its "waves" pass a point
every second.

--  Whatever you say is the speed of the pendulum, that speed
can only be true at one or two points in the pendulum's swing,
and it's different everywhere else in the swing.

-- The frequency of a pendulum depends only on the length
of the string from which it hangs.


If you take the given information and try to apply wave motion to it:

             Wave speed = (wavelength) x (frequency)

             Frequency  =  (speed) / (wavelength) ,

you would end up with

             Frequency = (30 meter/sec) / (0.35 meter) = 85.7 Hz

Have you ever seen anything that could be described as
a pendulum, swinging or even wiggling back and forth
85 times every second ? ! ?     That's pretty absurd. 

This math is not applicable to the pendulum.

6 0
3 years ago
What is the thickness of the disk of the milky way?
Temka [501]
its average thickness is 10,000 light-years,
3 0
4 years ago
A 120.0 kg crate is placed on a 15.00°
Citrus2011 [14]

F = 2820.1 N

Explanation:

Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as

Fnet = ma = 0 (a = 0 no sliding)

= F - mgsin15°

= 0

or

F = mgsin15°

= (120 kg)(9.8 m/s^2)sin15°

= 2820.1 N

7 0
3 years ago
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