Possibly, if you have list of densities and you have to match it. I can't think of any other scenarios in which it would be able to.
Hope I helped! :)
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
=>
=>
Answer:
Current = 8696 A
Fraction of power lost = = 0.151
Explanation:
Electric power is given by
where I is the current and V is the voltage.
Using values from the question,
The power loss is given by
where R is the resistance of the wire. From the question, the wire has a resistance of per km. Since resistance is proportional to length, the resistance of the wire is
Hence,
The fraction lost =
Answer:
The ice melts mass is:
Explanation:
Kinetic Energy
Heat gained by ice= mass(g) x 80 cal
( 1 cal = 4.184 *10^7er or g cm^2/ sec^2)
Assuming no loss in heat, in the motion so both continue with temperature 0~C
To find so the mass (gm) of ice melted
Answer:
83%
Explanation:
On the surface, the weight is:
W = GMm / R²
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.
In orbit, the weight is:
w = GMm / (R+h)²
where h is the height of the shuttle above the surface of the Earth.
The ratio is:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
The shuttle in orbit retains 83% of its weight on Earth.