An underground tunnel has two openings, with one opening a few meters higher than the other. If air moves past the higher openin
1 answer:
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Answer:
Explanation:
When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:
Where is the length of the string and
the velocity of propagation. Use this expression to find the value of
.
The velocity of propagation is given by the expression:
Where is the desirable variable of the problem, the linear mass density, and
is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:
With the value of the tension and the velocity you can find the mass density:
Answer:
The torque about his shoulder is 34.3Nm.
The solution approach assumes that the weight of the boy's arm acts at the center of the boy's arm length 35cm from the shoulder.
Explanation:
The solution to the problem can be found in the attachment below.
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.