A cubical surface surrounds a point charge q . Describe what happens to the total flux through the surface if (d) the charge is
1 answer:
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Answer:
Velocity of the ball after 3.04 (s) = 29.79 (m/s)
Explanation:
From the free fall movement we have the following formulas: and
, First we need to find the height to time iqual to 3.04 s using the formula:
and remember that golf ball was released from the rest (Vo= 0 (m/s)) so
, we get: h = -45.28 (m) with the height that we have got, now the velocity of the ball is calculate using
solving for Vf, we get:
replacing the values given
, so we get: Vf = 29.79 (m/s).
This question involves the concepts of the law of conservation of momentum.
The magnitude of the final momentum of the eight ball is "0.22 N.s".
According to the law of conservation of momentum:
where,
= initial momentum of the cue ball = 0.23 N.s
= initial momentum of the eight ball = 0 N.s (since ball is initially at rest)
= final momentum of the cue ball = 0.01 N.s
= final momentum of the eight ball = ?
Therefore,
Learn more about the law of conservation of momentum here:
Answer
Pressure, P = 1 atm
air density, ρ = 1.3 kg/m³
a) height of the atmosphere when the density is constant
Pressure at sea level = 1 atm = 101300 Pa
we know
P = ρ g h
h = 7951.33 m
height of the atmosphere will be equal to 7951.33 m
b) when air density decreased linearly to zero.
at x = 0 air density = 0
at x= h ρ_l = ρ_sl
assuming density is zero at x - distance
now, Pressure at depth x
integrating both side
now,
h = 15902.67 m
height of the atmosphere is equal to 15902.67 m.