Answer:
i. Cv =3R/2
ii. Cp = 5R/2
Explanation:
i. Cv = Molar heat capacity at constant volume
Since the internal energy of the ideal monoatomic gas is U = 3/2RT and Cv = dU/dT
Differentiating U with respect to T, we have
= d(3/2RT)/dT
= 3R/2
ii. Cp - Molar heat capacity at constant pressure
Cp = Cv + R
substituting Cv into the equation, we have
Cp = 3R/2 + R
taking L.C.M
Cp = (3R + 2R)/2
Cp = 5R/2
I would say B. Because actual mass would ricochet off the sidewalk.
250kg
would have momentum that is being caried by the impact of the trow
Answer: B. Concrete
Explanation:
Let N = reacting force pressing the bodies in context together (units in Newtons),
The question stated that the force pressing the two mounted/stacked objects together is equal to the weight of the object on top.
We need to start by finding the weight of the piece of wood.
friction is given by
f = μN
The value of f is 22.5,
and from the chart reference the coefficient of friction between wood and stone, μ is 0.30.
22.5 = 75. 0.30
Putting the values into the equation: 22.5 = 0.30N.
Divide both sides by 0.30 to find the value of N:
N= 22.5/0.3 = 75
Now that the piece of wood will be placed on another surface, its weight of 75 Newton is the force pressing the two bodies together.
To determine the new surface, you should find the new coefficient of friction by using the new value of the force of friction given 46.5:
46.5 = µ(75).
Divide both sides by 75 to isolate μ.
The refer chart also indicates that the coefficient of friction equals 0.62 between wood and concrete, so the new surface corresponding to 0.62 is the concrete, which is (B).
To develop this problem it is necessary to apply the concepts related to a magnetic field in spheres.
By definition we know that the magnetic field in a sphere can be described as

Where,
a = Radius
z = Distance to the magnetic field
I = Current
Permeability constant in free space
Our values are given as
diameter of the sphere then,

Thus z = a



Re-arrange to find I,



Therefore the current at the pole of this sphere is 