I think that the answer to that is true hope that helps
Answer:
844°C
Explanation:
The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.
We know that Fick's law is given by,

Where
is the concentration of gradient
D is the diffusivity coefficient
and J is the flux of atoms.
In the other hand we have, that

Where
is the proportionality constant,
R is the gas constant, T the temperature and
is the activation energy.
Replacing the value of diffusivity coefficient in Fick's law we have,

Rearrange the equation to get the value of temperature,

We have all the values in our equation.






Substituting,



Answer:
0.34 m
Explanation:
From the question,
v = λf................ Equation 1
Where v = speed of sound, f = frequency, λ = Wave length
Make λ the subject of the equation
λ = v/f............... Equation 2
Given: v = 340 m/s, f = 500 Hz.
Substitute these values into equation 2
λ = 340/500
λ = 0.68 m
But, the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length
Therefore,
λ/2 = 0.68/2
λ/2 = 0.34 m
Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m
Answer:
2632 foot-pound
Explanation:
Work done: Work is said to be done when ever a force moves a body through a given distance. The S.I unit of force is Newton (N).
From the question,
The expression for work done is given as,
W = Fdcos∅......................... Equation 1
Where W = work done, F = force, d = distance, ∅ = angle between the force and the horizontal.
Given: F = 32 lbs, d = 90 feet, ∅ = 24°
substitute into equation 1
W = 32×90×cos24
W = 2880(0.914)
W = 2632.32
W = 2632 foot-pound
Answer:
True
Explanation:
When a nuclear power plants ends its operating life, it is necessary to dispose correctly all the radioactive material left from the operations of the power plant.
In fact, apart from the fuel rods (which contain the uranium that is the fuel of the nuclear reactions), other materials of the core (such as the vessel) becomes radioactive due to the prolonged exposure to the products of nuclear reactions in the core.
As a result, all these materials remain radioactive for very long time (hundreds or thousands of years). It is therefore important to dispose them correctly, in proper sealed containers which are able to shield the radiation emitted by these radioactive element (alpha, beta and gamma radiation).
Therefore, "waste from nuclear power plants must be disposed of in radioactively shielded storage containers" is a correct statement.