Learning Goal: To practice Problem-Solving Strategy 19.1 Work in Ideal-gas Processes. A cylinder with initial volume VVV contain
s a sample of a gas at pressure ppp. On one end of the cylinder, a piston is let free to move so that the gas slowly expands in such a way that its pressure is directly proportional to its volume. After the gas reaches the volume 3V3V and pressure 3p3p, the piston is pushed in so that the gas is compressed isobarically to its original volume VVV. The gas is then cooled isochorically until it returns to the original volume and pressure. Find the work WWW done on the gas during the entire process.
1 answer:
You might be interested in
Answer:
<em>a) 318.2 W/m^2</em>
<em>b) 2.5 x 10^-4 J</em>
<em>c) 1.55 x 10^-8 v/m</em>
<em></em>
Explanation:
Power of laser P = 1 mW = 1 x 10^-3 W
exposure time t = 250 ms = 250 x 10^-3 s
If beam diameter = 2 mm = 2 x 10^-3 m
then
cross-sectional area of beam A = = (3.142 x
)/4
A = 3.142 x 10^-6 m^2
a) Intensity I = P/A
where P is the power of the laser
A is the cros-sectional area of the beam
I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>
<em></em>
b) Total energy delivered E = Pt
where P is the power of the beam
t is the exposure time
E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>
<em></em>
c) The peak electric field is given as
E =
where I is the intensity of the beam
E is the electric field
c is the speed of light = 3 x 10^8 m/s
= 8.85 x 10^9 m kg s^-2 A^-2
E = = <em>1.55 x 10^-8 v/m</em>
Answer : Tension in the line = 936.7 N
Explanation :
It is given that,
Mass of student, m = 65 kg
The angle between slackline and horizontal,
The two forces that acts are :
(i) Tension
(ii) Weight
So, from the figure it is clear that :
Hence, this is the required solution.
