Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2 ![v_{f}](https://tex.z-dn.net/?f=v_%7Bf%7D)
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²
Answer:
The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame
Explanation:
From the question we are told that
The distance between earth and Retah is ![d = 20 \ light \ hours = 20 * 3600 * c = 72000c \ m](https://tex.z-dn.net/?f=d%20%3D%2020%20%5C%20light%20%5C%20hours%20%3D%20%2020%20%2A%203600%20%2A%20%20c%20%3D%20%2072000c%20%5C%20m)
Here c is the peed of light with value ![c = 3.0*10^8 m/s](https://tex.z-dn.net/?f=c%20%3D%20%203.0%2A10%5E8%20m%2Fs)
The time taken to reach Retah from earth is ![t = 25 \ hours = 25 * 3600 =90000 \ sec](https://tex.z-dn.net/?f=t%20%3D%20%2025%20%5C%20hours%20%20%3D%20%2025%20%2A%203600%20%3D90000%20%5C%20sec)
The velocity of the spacecraft is mathematically evaluated as
![v_s = \frac{d }{t}](https://tex.z-dn.net/?f=v_s%20%3D%20%20%5Cfrac%7Bd%20%7D%7Bt%7D)
substituting values
![v_s = \frac{72000 * 3.0*10^{8} }{90000}](https://tex.z-dn.net/?f=v_s%20%3D%20%20%5Cfrac%7B72000%20%2A%203.0%2A10%5E%7B8%7D%20%7D%7B90000%7D)
![v_s = 2.40*10^{8} \ m/s](https://tex.z-dn.net/?f=v_s%20%3D%20%202.40%2A10%5E%7B8%7D%20%5C%20m%2Fs)
The time elapsed in the spacecraft’s frame is mathematically evaluated as
![T = t * \sqrt{ 1 - \frac{v^2}{c^2} }](https://tex.z-dn.net/?f=T%20%20%3D%20%20t%20%2A%20%20%5Csqrt%7B%201%20-%20%20%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%20%7D)
substituting value
![T = 90000 * \sqrt{ 1 - \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }](https://tex.z-dn.net/?f=T%20%20%3D%20%2090000%20%2A%20%20%5Csqrt%7B%201%20-%20%20%5Cfrac%7B%5B2.4%2A10%5E%7B8%7D%5D%5E2%7D%7B%5B3.0%2A10%5E%7B8%7D%5D%5E2%7D%20%7D)
![T = 54000 \ s](https://tex.z-dn.net/?f=T%20%3D%2054000%20%5C%20s)
=> ![T = 15 \ hours](https://tex.z-dn.net/?f=T%20%20%3D%2015%20%5C%20hours)
So The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame
Answer:
a force
Explanation:
a force causes a certain object to move and make a displacement.
Answer:Both are correct
Explanation:
Both are correct because
Mechanical efficiency is the dimensionless term which is the ratio of brake horsepower to the Indicated horse Power
Where brake power is the Power obtained at the crankshaft and
Indicated horsepower is the power obtained in the combustion chamber and this power is the loss in the form of friction.
Volumetric efficiency is the ratio of actual fuel intake to the maximum air fuel that could be taken.
your answer would be The one in the top right corner that looks a bit like this..