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2 answers:
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Answer:
i) The motion of both particles are shown on the same speed-time curve included
ii) Approximately 19.5 seconds
Explanation:
We are given that;
Initial velocity of particle, P = 140 m/s
Start time of particle P = 6 s before start time of particle Q
Position of particle Q when velocity is 25 m/s = 125 m
Therefore, from the equation of motion, we have for particle Q;
v² = u² + 2·a·s
Where:
v = Final velocity = 25 m/s
u = Initial velocity = 0 m/s
a = Acceleration
s = Distance covered = 125 m
Therefore;
25² = 0² + 2×a×125
Which gives a = 25²/(2×125) = 2.5 m/s²
The time taken for particle Q to reach 125 m is found from the relation;
s = u·t + 1/2·a·t²
Where:
t = Time of journey
Therefore;
125 = 0×t + 1/2×2.5×t²
Which gives 125 = 1.25 × t²
Hence, t² = 125/1.25 = 100
t = √(100) = 10 s
The equation for particle Q is v = 0 + 2.5×t
Hence, since particle P starts deceleration 6 seconds before the commencement of motion of particle Q, the amount of seconds after the commencement of deceleration of the first particle P that it takes for particle P to come to rest is found as follows;
Hence, at t = 6 + 10 = 16 seconds particle P speed = 25 m/s
From the equation of motion, for particle P (decelerating) we have
v = u - a·t
Where:
v = 25 m/s
u = 140 m/s
t = 16 s
Hence, 25 = 140 - a×16
∴ 16·a = 140 - 25 = 115
a = 115/16 = 7.1875 m/s²
Therefore, the time it takes before particle P comes to rest is found from the same equation of motion, where v = 0 as follows;
v = u - a·t
0 = 140 - 7.1875 × t
∴7.1875·t = 140
t = 140/7.1875 = 19.48 s ≈ 19.5 seconds.
Answer:
Part a)
Part b)
this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.
Part C)
here we can assume the sphere is placed at vacuum so that there is no break down of air.
Explanation:
Part a)
As we know that the potential near the surface of metal sphere is given by the equation
here we have
Q = 8 C
R = 10.0 cm
now we have
Part b)
this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.
Part C)
here we can assume the sphere is placed at vacuum so that there is no break down of air.