A 0.50-kg object moves on a horizontal frictionless circular track with a radius of 2.5 m. An external force of 3.0 N, always ta
1 answer:
Answer:
Radial force component of force = 37.68 N
Explanation:
By Newton's 2 nd law of motion,
F = ma
F = 3.0 N, m = 0.5 Kg, a (Linear Acceleration ) = ?
3 = 0.5 a
a = 6
Now, a = 6 , r = 2.5 m ,α(angular acceleration ) = ?
a = r α
6= 2.5 α
hence α = 2.4 rad/
we know that,
Radial force component of force =
since m =0.5 kg, r =2.5 m w =5.49 rad/sec
Radial force component of force = = 37.68 N
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Hence the work by the body W = FS
If the force is not along the direction of displacement,then the work by a body for travelling a distance S will be -
where
is the component of the force along the direction of displacement.
As per the question the power P is given as -
Hence alternative definition of power P = F.V
Answer:
- The work made by the gas is 7475.69 joules
- The heat absorbed is 7475.69 joules
Explanation:
<h3>Work</h3>We know that the differential work made by the gas its defined as:
We can solve this by integration:
but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law
This give us
As n, R and T are constants
But the volume is:
Now, lets use the value from the problem.
The temperature its:
The ideal gas constant:
So:
We know that, for an ideal gas, the energy is:
where its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.
By the first law of thermodynamics, we know
where is the Work made by the gas (please, be careful with this sign convention, its not always the same.)
So: