How much potential energy does a 40-kg medicine ball gain when it is lifted 5 m? Use the formula: GPE= mass X gravity (9.8 m/s2)
1 answer:
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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.
Answer:
d = 185.26 meters
Explanation:
It is given that,
Launching angle of the projectile,
Initial speed of the projectile, u = 48 m/s
Let at distance d the projectile hits the ground from the launch point. It is equal to range of the projectile. Its formula is given by :
Substituting all the values in above formula. So, we get :
d = 185.26 meters
So, the distance between the launch point and the point where it hits is 185.26 meters. Hence, this is the required solution.
Answer:
31.831 Hz.
Explanation:
<u>Given:</u>
The vertical displacement of a wave is given in generalized form as
<em>where</em>,
- A = amplitude of the displacement of the wave.
- k = wave number of the wave =
= wavelength of the wave.
- x = horizontal displacement of the wave.
= angular frequency of the wave =
.
- f = frequency of the wave.
- t = time at which the displacement is calculated.
On comparing the generalized equation with the given equation of the displacement of the wave, we get,
therefore,
It is the required frequency of the wave.