What is the pressure exerted when a force of 20 N is applied to an object over an area of 2 m2 ?
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Answer:
6.57 m/s
Explanation:
First use Hook's Law to determine the F the compressed spring acts on the mass. Hook's Law F=kx; F=force, k=stiffnes of spring (or spring constant), x=displacement
F=kx; F=180(.3) = 54 N
Next from Newton's second law find the acceleration of the mass.
Newton's .2nd law F=ma; a=F/m ; a=54/.75 = 72m/s²
Now use the kinematic equation for velocity (or speed)
v₂²= v₀² + 2a(x₂-x₀); v₂=final velocity; v₀=initial velocity; a=acceleration; x₂=final displacement; x₀=initial displacment.
v₀=0, since the mass is at rest before we release it
a=72 m/s² (from above)
x₀=0 as the start position already compressed
x₂=0.3m (this puts the spring back to it's natural length)
v₂²= 0 + 2(72)(0.3) = 43.2 m²/s²
v₂= = 6.57 m/s
The mass of water that must be raised is
Explanation:
Since the process is 70% efficiency, the power in output to the turbine can be written as
where is the power in input.
The power in input can be written as
where
W is the work done in lifting the water
t = 3 h = 10,800 s is the time elapsed
The work done in lifting the water is given by
where
m is the mass of water
is the acceleration of gravity
h = 45 m is the height at which the water is lifted
Combining the three equations together, we get:
Where
And solving for m, we find:
Learn more about power:
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