Two balls of equal mass at the bottom of an incline are rolled upward without slipping at the same initial velocity. one ball is
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G = 9.81 m/sec^2) g = 9.81
<span>Solving for velocity : </span>
<span> = 2gh </span>
<span>v = </span>
<span>v = (2 x 9.81 x 10)^1/2 </span>
<span>v = 196.2 m/sec (answer)</span>
<span>Solving for velocity : </span>
<span>v = </span>
<span>v = (2 x 9.81 x 10)^1/2 </span>
<span>v = 196.2 m/sec (answer)</span>
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.