A small sphere of mass m is launched horizontally over a body of water from a height h above the water and with a launch speed v
1 answer:
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The initial volume of the gas is
while its final volume is
so its variation of volume is
The pressure is constant, and it is
Therefore the work done by the gas is
where the negative sign means the work is done by the surrounding on the gas.
The heat energy given to the gas is
And the change in internal energy of the gas can be found by using the first law of thermodynamics:
where the positive sign means the internal energy of the gas has increased.
while its final volume is
so its variation of volume is
The pressure is constant, and it is
Therefore the work done by the gas is
where the negative sign means the work is done by the surrounding on the gas.
The heat energy given to the gas is
And the change in internal energy of the gas can be found by using the first law of thermodynamics:
where the positive sign means the internal energy of the gas has increased.
To solve this problem it is necessary to apply the concepts related to the heat flux rate expressed in energetic terms. The rate of heat flow is the amount of heat that is transferred per unit of time in some material. Mathematically it can be expressed as:
Where
k = 0.84 J/s⋅m⋅°C (The thermal conductivity of the material)
Area
Length
= Temperature of the "hot"reservoir
= Temperature of the "cold"reservoir
Replacing with our values we have that,
Therefore the correct answer is B.
Answer:
0.5% per oscillation
Explanation:
The term 'damped oscillation' means an oscillation that fades away with time. For Example; a swinging pendulum.
Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).
Where m= Mass, v= velocity.
Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).
Where k= force constant, X= displacement.
Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).
Therefore, we use equation (3) to get the resonance frequency,
W^2= k/m--------------------------------------------------------------------------------------(3)
Slotting values into equation (3).
= 10/2.5.
= ✓4.
= 2 s^-1.
Recall that, F= -kX
F^2= (-0.1)^2
Potential energy,PE= 1/2 ×0.01
Potential energy= 0.05 ×100
= 0.5% per oscillation.
