<span>The work output of a machine divided by the work input is the "Efficiency" of the machine.
Hope this helps!</span>
Answer:
the frequency of the second harmonic of the pipe is 425 Hz
Explanation:
Given;
length of the open pipe, L = 0.8 m
velocity of sound, v = 340 m/s
The wavelength of the second harmonic is calculated as follows;
L = A ---> N + N--->N + N--->A
where;
L is the length of the pipe in the second harmonic
A represents antinode of the wave
N represents the node of the wave

The frequency is calculated as follows;

Therefore, the frequency of the second harmonic of the pipe is 425 Hz.
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 
Here c is the peed of light with value 
The time taken to reach Retah from earth is 
The velocity of the spacecraft is mathematically evaluated as

substituting values


The time elapsed in the spacecraft’s frame is mathematically evaluated as

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)

=> 
So The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame
Answer: 704
Explanation:Vi = 0 m/s
vf = 65 m/s
a = 3 m/s2
d = ??
vf2 = vi2 + 2*a*d
(65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d
4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d
(4225 m 2/m2)/(6 m/s2) = d
d = 704 m