Answer: 45000 m/s^2
Explanation:
First, we know:
Bullet's Initial Velocity (Vo) = 0 m/s
Distance Travelled (D) = 1 m
Final Velocity (Vf) = 300 m/s
We need to know the bullets acceleration to reach the final velocity on that distance. Because the bullet stars from rest, his Initial Velocity is equal to zero.
We dont know on how much time it takes for the bullet, but we know the acceleration on the barrel is constant.
We can use the following equation for this
Because Vo is zero, the equation simplify, so we have.
Searching for the accelaration we obtain
Changing values
This is the acceleration needed to reach the velocity at the end of the barrel.
Answer:E. an object will remain in uniform motion unless acted upon by a force
Explanation:
Answer:
Assume that this wavelength is measured in vacuum. The energy on each photon of this wave would be approximately .
Explanation:
The Planck-Einstein Relation relates the energy of a photon to its frequency :
,
where is Planck's Constant.
.
This question did not provide the frequency of this wave directly; the value of needs to be calculated from the wavelength of this wave. Assume that this wave is travelling at the speed of light in vacuum:
.
The frequency of this electromagnetic wave would be:
.
Apply the Planck-Einstein Relation to find the energy of a photon of this electromagnetic wave:
.
Note that combining the two equations above ( and ) will give:
.
This equation is supposed to give the same result (energy of a photon of this wave given its wavelength and speed) in one step:
.
I assume you're talking about a pilot. If the ejection seat has an acceleration of 8<em>g</em>, then it would exert a normal force of 8<em>g</em> (70 kg) ≈ 5600 N.
(This is assuming the pilot is flying horizontally at a constant speed, and the seat is ejected vertically upward.)
To reiterate, this is *only* the force exerted by the seat on the pilot. Contrast this with the <em>net</em> force on the pilot, which would be the normal force minus the pilot's weight, 5600 N - (70 kg)<em>g</em> ≈ 4900 N.
If instead the seat ejects the pilot directly downward, the force exerted by the seat would have the same magnitude of 5600 N, but its direction would be reversed to point downward, making it negative. But the <em>net</em> force would change to -5600 N - (70 kg)<em>g</em> ≈ -6300 N