The average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.
<h3>Acceleration of the box</h3>
The acceleration of the box is calculated as follows;
vf² = vi² + 2as
a = (vf² - vi²)/2s
a = (11.5² - 13²) / (2 x 8.5)
a = -2.16 m/s²
<h3>Time of motion of the box</h3>
The time taken for the box to travel is calculated as follows;
a = (vf - vi)/t
t = (vf - vi) / a
t = (11.5 - 13) / (-2.16)
t = 0.69 s
<h3>Average power supplied by the friction</h3>
P = Fv
P = (ma)(vf - vi)
P = (1 x -2.16) x (11.5 - 13)
P = 3.24 W
Thus, the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.
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Answer:
91.84 m/s²
Explanation:
velocity, v = 600 m/s
acceleration, a = 4 g = 4 x 9.8 = 39.2 m/s^2
Let the radius of the loop is r.
he experiences a centripetal force.
centripetal acceleration,
a = v² / r
39.2 x r = 600 x 600
r = 3600 / 39.2
r = 91.84 m/s²
Thus, the radius of the loop is 91.84 m/s².
The first resonant position below the open end of the resonance tube is; <em><u>one-quarter of the wavelength</u></em>
In the event of the first resonant position in a resonance tube, there will be a maximum air displacement which is only one antinode right at the open end where the motion is constrained.
However, there will be no displacement at the closed end which means another one node right at the closed end where air is halted.
This means that the standing wave will have one-quarter of the wavelength in the test tube.
Thus;
L = ¼λ
Read more at; brainly.com/question/17086525
To calculate that, we'd need to know the coefficient of static friction between the block and the surface, which you haven't told us.
This one is correct
Jamie is correct, because the mechanical energy is converting to electrical energy.
