How much work in J does the string do on the boy if the boy stands still?
<span>answer: None. The equation for work is W = force x distance. Since the boy isn't moving, the distance is zero. Anything times zero is zero </span>
<span>--------------------------------------... </span>
<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite? </span>
<span>answer: might be a trick question since his direction away from the kite and his velocity weren't noted. Perhaps he just set the string down and walked away 11m from the kite. If he did this, it is the same as the first one...no work was done by the sting on the boy. </span>
<span>If he did walk backwards with no velocity indicated, and held the string and it stayed at 30 deg the answer would be: </span>
<span>4.5N + (boys negative acceleration * mass) = total force1 </span>
<span>work = total force1 x 11 meters </span>
<span>--------------------------------------... </span>
<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m toward the kite? </span>
<span>answer: same as above only reversed: </span>
<span>4.5N - (boys negative acceleration * mass) = total force2 </span>
<span>work = total force2 x 11 meters</span>
Answer: Escaped volume = 0.0612m^3
Explanation:
According to Boyle's law
P1V1 = P2V2
P1 = initial pressure in the tire = 36.0psi + 14.696psi = 50.696psi (guage + atmospheric pressure)
P2 = atmospheric pressure= 14.696psi
V1 = volume of tire =0.025m^3
V2 = escaped volume + V1 ( since air still remain in the tire)
V2 = P1V1/P2
V2 = 50.696×0.025/14.696
V2 = 0.0862m^3
Escaped volume = 0.0862 - 0.025 = 0.0612m^3
Answer: die
Explanation: oyxagan all goon bc of all dat suffs
Answer:
1070 Hz
Explanation:
First, I should point out there might be a typo in the question or the question has inconsistent values. If the tube is 40 cm long, standing waves cannot be produced at 42.5 cm and 58.5 cm lengths. I assume the length is more than the value in the question then. Under this assumption, we proceed as below:
The insert in the tube creates a closed pipe with one end open and the other closed. For a closed pipe, the difference between successive resonances is a half wavelength 
.
Hence, we have
.
The speed of a wave is the product of its wavelength and its frequency.
