Answer:
Explanation:
10. As a spaceship is moving toward Earth, an Earthling measures its length to be 325 m, while the captain on board radios that her spaceship's length is 1150 m. (c = 3.00 × 108 m/s) (a) How fast is the rocket moving relative to Earth? (b) What is the TOTAL energy of a 75.0-kg crewman as measured by (i) the captain in the rocket and (ii) the Earthling?
To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of
(Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as

Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing

![W = (14*7*62)\big [15y-\frac{y^2}{2}\big ]^{15}_0](https://tex.z-dn.net/?f=W%20%3D%20%2814%2A7%2A62%29%5Cbig%20%5B15y-%5Cfrac%7By%5E2%7D%7B2%7D%5Cbig%20%5D%5E%7B15%7D_0)
![W = (14*7*62)[15(15)-\frac{(15)^2}{2}]](https://tex.z-dn.net/?f=W%20%3D%20%2814%2A7%2A62%29%5B15%2815%29-%5Cfrac%7B%2815%29%5E2%7D%7B2%7D%5D)

Therefore the total work in the system is 
Answer:
Explanation:
The 2 equations we need here are, first:
and then once we solve for the acceleration here:
Δx
Solving for acceleration:
and now we will use that in the other equation:
Δx and
36 = 16 +
Δx and
20 =
Δx and
Δx so
Δx = 50 m
F = ma
<span>where </span>
<span>F = frictional force </span>
<span>m = mass of the block = 1.4 kg (given) </span>
<span>a = acceleration of the block = 1.25 m/sec^2 (given) </span>
<span>Substituting values, </span>
<span>F = (1.4)(1.25) </span>
<span>F = 1.75 N </span>
<span>By definition, </span>
<span>F = (mu)(Normal force) </span>
<span>where </span>
<span>mu = coefficient of friction </span>
<span>Normal force = mg = 1.4*9.8 = 13.72 </span>
<span>Again, substituting appropriate values, </span>
<span>1.75 = mu(13.72) </span>
<span>mu = 0.128</span>
Answer: The Propellant fraction is 0.87.
The payload fraction is 0.04.
Δv = 8991.81 m/s
Explanation: To determine the fractions, first, calculate the total mass of the rocket:



The Propellant Fraction will be


0.87
The Payload Fraction is:


0.04
The value of Δv is calculated by the formula:
Δv = 
The exhaust velocity (
) is:

9.81*450
4414.5
is the total mass after the rocket consume all the propellant and
is the total mass before the action.
Δv = 
Δv = 
Δv = - 4414.5.ln(0.13)
Δv = 8991.81
Δv will be 8991.81 m/s.