Answer:
During an earthquake, seismic waves travels through the Earth's interior as body or p waves.
Explanation:
If neither of the bold words look familiar from your lesson feel free to ignore this answer
Answer:
a) w = 7.27 * 10^-5 rad/s
b) v1 = 463.1 m/s
c) v1 = 440.433 m/s
Explanation:
Given:-
- The radius of the earth, R = 6.37 * 10 ^6 m
- The time period for 1 revolution T = 24 hrs
Find:
What is the earth's angular speed?
What is the speed of a point on the equator?
What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?
Solution:
- The angular speed w of the earth can be related with the Time period T of the earth revolution by:
w = 2π / T
w = 2π / 24*3600
w = 7.27 * 10^-5 rad/s
- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.
v1 = R*w
v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)
v1 = 463.1 m/s
- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
π/2 ........... s
x ............ 1/5 s
x = π/2*5 = 18°
- The radius of the earth R' at point where θ = 18° from the equator is:
R' = R*cos(18)
R' = (6.37 * 10 ^6)*cos(18)
R' = 6058230.0088 m
- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.
v2 = R'*w
v2 = (6058230.0088)*(7.27 * 10^-5)
v2 = 440.433 m/s
Answer:
The maximum amount of work is 
Explanation:
From the question we are told that
The temperature of the environment is 
The volume of container A is 
Initially the number of moles is 
The volume of container B is 
At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as
![W = P_A V_A ln[ \frac{V_B}{V_A} ]](https://tex.z-dn.net/?f=W%20%3D%20%20P_A%20V_A%20%20ln%5B%20%5Cfrac%7BV_B%7D%7BV_A%7D%20%5D)
Now from the Ideal gas law
So substituting for 
in the equation above
![W = nRT ln [\frac{V_B}{V_A} ]](https://tex.z-dn.net/?f=W%20%3D%20%20nRT%20ln%20%5B%5Cfrac%7BV_B%7D%7BV_A%7D%20%5D)
Where R is the gas constant with a values of 
Substituting values we have that
![W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]](https://tex.z-dn.net/?f=W%20%3D%201.2%20%2A%20%288.314%29%20%2A%20%28280%29%20%2A%20ln%20%5B%5Cfrac%7B3.5%7D%7B2%7D%20%5D)
