At the surface of Venus the average temperature is a balmy 460∘C due to the greenhouse effect (global warming!), the pressure is
92 Earth-atmospheres, and the acceleration due to gravity is 0.894 gEarthgEarth. The atmosphere is nearly all CO2 (molar mass 44.0g/m) and the temperature remains remarkably constant. We shall assume that the temperature does not change at all with altitude.
a. What is the atmospheric pressure of 2.00 km above the surface of Venus? Express your answer in Earth-atmospheres.
b. What is the atmospheric pressure of 2.00 km above the surface of Venus? Express your answer in Venus-atmospheres.
c. What is the root-mean-square speed of the CO2 molecules at the surface of Venus and at an altitude of 2.00 km?
a. What is the atmospheric pressure of 2.00 km above the surface of Venus? Express your answer in Earth-atmospheres.
b. What is the atmospheric pressure of 2.00 km above the surface of Venus? Express your answer in Venus-atmospheres.
c. What is the root-mean-square speed of the CO2 molecules at the surface of Venus and at an altitude of 2.00 km?
1 answer:
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Answer:
The increase in the gravitational potential energy is 29.93 joules.
Explanation:
Given that,
Mass of the box, m = 2.35 kg
It is lifted from the floor to a tabletop 1.30 m above the floor, h = 1.3 m
We need to find the increase the gravitational potential energy. Initial it will placed at ground i.e. its initial gravitational potential is equal to 0. The increase in the gravitational potential energy is given by :
U = 29.93 Joules
So, the increase in the gravitational potential energy is 29.93 joules. Hence, this is the required solution.
Complete Question
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.
Answer:
The torque is
Explanation:
From the question we are told that
The mass of the steel ball is
The length of arm is
The mass of the arm is
Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as
=>
=>
Generally the magnitude of torque about the athlete shoulder is mathematically represented as
=>
=>
Here We can use principle of angular momentum conservation
Here as we know boy + projected mass system has no external torque
Since there is no torque so we can say the angular momentum is conserved
now we know that
m = 2 kg
v = 2.5 m/s
L = 0.35 m
I = 4.5 kg-m^2
now plug in all values in above equation
so the final angular speed will be 0.37 rad/s