0.5kg ball moving at a speed of 3m/s rolls up a hill what is thet height
1 answer:
Answer:
0.459 m.
Explanation:
To calculate the height, use 3rd equation of motion as
Here v is the final velocity, u is the initial velocity, g is the acceleration due to gravity and h is the height.
As ball moving up so acceleration due to gravity becomes negative i.e -g and at maximum height v = 0.
Therefore equation becomes,
Given u = 3 m/s and g = 9.8 m/s^2.
Substitute the given values, we get
h = 0.459 m
Thus, the height of hill is 0.459 m.
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Answer:
The wavelength of the light decreases as it enters into the medium with the greater optical density.
The frequency of the light remains constant as it transitions between materials.
Explanation:
- When a ray of light crosses a boundary between two different materials, it undergoes refraction: the ray changes direction, and it also changes speed, according to the relationship:
where v is the speed of the ray of light in the material, c is the speed of light in a vacuum, n is the index of refraction of the material, which is larger for a medium with larger optical density. So, from the equation, we see that the larger the optical density, the smaller the speed of the wave.
- The frequency of the light does not depend on the properties of the medium, so it remains unchanged: therefore the statement
The frequency of the light remains constant as it transitions between materials.
is correct.
- Moreover, the wavelength of the ray of light is related to the speed and the frequency by the equation
where v is the speed and f the frequency. Since we have seen that v decreases and f remains constant, this means that the wavelength decreases as well, so the statement
The wavelength of the light decreases as it enters into the medium with the greater optical density.
is also correct.