In what do electromagnetic waves cause disturbances when they transfer
2 answers:
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Answer:
The work done is 3.4 × 10⁵ J.
Explanation:
Given:
Pressure of the gas produced (P) = 179 kPa
Volume of the gas produced (ΔV) = 1.90 m³
We need to find the work done in joules. For that, we don't need any conversion as the units are already in SI units which will give the result in Joules only.
Now, let us verify our results by using conversion factors and without using them.
Using conversion factors:
1 m³ = 1000 L
So, 1.90 m³ = 1.90 m³ × 1000 = 1900 L
Also, 1 atm = 101.325 kPa
So, 179 kPa = 179 kPa × = 1.767 atm
Now, work done in a constant pressure process is given as:
Work = Pressure × Volume change
Work = P × ΔV
Work = 1.767 atm × 1900 L
Work = 3.36 × 10³ atm-L
Now, again using the energy conversion for work.
1 atm-L = 101.325 J
So, 3.36 × 10³ atm-L = 3.36 × 10³ atm-L × = 3.4 × 10⁵ J
Therefore, the work done is 3.4 × 10⁵ J.
Now, let us verify the above result without any conversion.
Work = P × ΔV = 179 × 1000 × 1.90 = 3.4 × 10⁵ J.
Therefore, the work done is same by both ways.
Hence the work done is 3.4 × 10⁵ J.
1) Electric potential inside the sphere:
2) Ratio Vcenter/Vsurface: 3/2
3) Find graph in attachment
Explanation:
1)
The electric field inside the sphere is given by
where
is the vacuum permittivity
Q is the charge on the sphere
R is the radius of the sphere
r is the distance from the centre at which we compute the field
For a radial field,
Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,
The potential at the surface, V(R), is that of a point charge, so
Therefore we can find the potential inside the sphere, V(r):
2)
At the center,
r = 0
Therefore the potential at the center of the sphere is:
On the other hand, the potential at the surface is
Therefore, the ratio V(center)/V(surface) is:
3)
The graph of V versus r can be found in attachment.
We observe the following:
- At r = 0, the value of the potential is , as found in part b) (where
)
- Between r and R, the potential decreases as
- Then at r = R, the potential is
- Between r = R and r = 3R, the potential decreases as , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (
)
Learn more about electric fields and potential:
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