A particle that is smaller than an atom or a cluster of particles.
Answer:
the claim is not valid or reasonable.
Explanation:
In order to test the claim we will find the maximum and actual efficiencies. maximum efficiency of a heat engine can be found as:
η(max) = 1 - T₁/T₂
where,
η(max) = maximum efficiency = ?
T₁ = Sink Temperature = 300 K
T₂ = Source Temperature = 400 K
Therefore,
η(max) = 1 - 300 K/400 K
η(max) = 0.25 = 25%
Now, we calculate the actual frequency of the engine:
η = W/Q
where,
W = Net Work = 250 KJ
Q = Heat Received = 750 KJ
Therefore,
η = 250 KJ/750 KJ
η = 0.333 = 33.3 %
η > η(max)
The actual efficiency of a heat engine can never be greater than its Carnot efficiency or the maximum efficiency.
<u>Therefore, the claim is not valid or reasonable.</u>
Answer: Increasing the frequency does not increase the wavelength. They are inversely related.
Explanation:
As wavelength increases, frequency decreases. If you look at a transverse wave and it has a long wavelength, there only a few waves produce. Which means there is less frequency produced. So as wavelength increases, frequency decreases. The other way around can work to. As frequency increases, wavelength decreases. They are inversely related.
The given question is incomplete. The complete question is as follows.
In a nuclear physics experiment, a proton (mass 
kg, charge +e = 
C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 
m/s. The proton comes momentarily to rest at a distance 
m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 
m apart?
Explanation:
The given data is as follows.
Mass of proton = 
kg
Charge of proton = 
Speed of proton = 
Distance traveled = 
We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.
= 
where, 
U = 
Putting the given values into the above formula as follows.
U =
= 
= 
Therefore, we can conclude that the electric potential energy of the proton and nucleus is .
