V^2-u^2=2as
v=final velocity=unkown
u=initial velocity=0 m/s, because freely falling
a=acceleration due to gravity=9.8 m/s^2
s=distance (here height) traveled=4.5m
therefore the final velocity,
v^2=2*9.8*4.5
v=<span>9.39m/s</span>
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Explanation:
Answer:
700 mL or 0.0007 m³
Explanation:
P₁ = Initial pressure = 2 atm
V₁ = Initial volume = 350 mL
P₂ = Final pressure = 1 atm
V₂ = Final volume
Here the temperature remains constant. So, Boyle's law can be applied here.
P₁V₁ = P₂V₂
So, volume of this sample of gas at standard atmospheric pressure would be 700 mL or 0.0007 m³
I think minerals can be both.
The magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s
The rotating equivalent of linear momentum in physics is called angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system stays constant—it is significant in physics. Both the direction and the amplitude of angular momentum are preserved.
Given the density of air to be 1.29 kg/m3 and a wind speed of 73.0 mi/h
We have to find the magnitude of the angular momentum
Let,
ρ = Density of air = 1.29 kg/m^3
v = Speed of wind = 73.0 mi/h = 0.032 km/s
M = angular momentum of air
Let the volume of air be 1 m^3
Mass = Volume x ρ = 1 x 1.29 = 1.29 kg
Momentum = M = mass x velocity
Momentum = 1.29 x 0.0032
Momentum = 4.128 x 10^(-3) kg·m^2/s
Hence the magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s
Learn more about angular momentum here:
brainly.com/question/7538238
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