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rusak2 [61]
3 years ago
5

TRUE OR FALSE Light can travel through materials that you cannot see through, like wood or aluminum foil.

Physics
1 answer:
Keith_Richards [23]3 years ago
8 0

i think its uhh......tralse

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Which other subatomic particle has the same mass as a neutron
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Answer:

.................... protons :)

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3 years ago
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If the amplitude of a wave is 4 feet, what is the Wave Height?
Alexus [3.1K]

A wave is a disturbance that moves along a medium from one end to the other. If one watches an ocean wave moving along the medium (the ocean water), one can observe that the crest of the wave is moving from one location to another over a given interval of time. The crest is observed to cover distance. The speed of an object refers to how fast an object is moving and is usually expressed as the distance traveled per time of travel. In the case of a wave, the speed is the distance traveled by a given point on the wave (such as a crest) in a given interval of time. In equation form,

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3 years ago
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A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the a
gizmo_the_mogwai [7]

Answer:

Acceleration,  767.08\ m/s^2

Explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

v^2=u^2+2ah

u = initial velocity=0 m/s

a = acceleration=g

v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 17.3} \\\\v=18.41\ m/s

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2

So, the average acceleration of the ball during the time it is in contact is 767.08\ m/s^2.

4 0
3 years ago
Three balls, with masses of 3m,2m and m, are fastened to a massless rod of length L. The rotational inertias about the ledt
Oduvanchick [21]

I = MR^2

The Attempt at a Solution:::

I total = (3M)(0)^2 + (2M)(L/2)^2 + (M)(L)^2

I total = 3ML^2/2

It says the answer is 3ML^2/4 though.

⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔

mark it as brainliest.... ✌✌✌

8 0
3 years ago
An air compressor compresses 6 L of air at 120 kPa and 22°C to 1000 kPa and 400°C. Determine the flow work, in kJ/kg, required b
Mariana [72]

Answer:

The work flow required by the compressor = 100.67Kj/kg

Explanation:

The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .

The work flow can be determined using the equation:

M1h1 + W = Mh2

U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2

Workflow = P2alpha2 - P1alpha1

Workflow = (h2 -U2) - (h1 - U1)

Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)

Workflow = ( 193.191 - 92.519)Kj/kg

Workflow = 100.672Kj/kg

6 0
3 years ago
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