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
=> 
=>
The velocity of shortening refers to the speed of the contraction from
the muscle shortening while lifting a load. The relationship between the
resistance and velocity of shortening is inverse. The greater the
resistance, the shorter the velocity of shortening and the smaller the
resistance, the larger the velocity of shortening.
Hopefully this help :)
A) 4.7 cm
The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is
where
n is the order of the maximum
is the wavelength
is the distance between the slits
In this problem,
n = 5
So we find
And given the distance of the screen from the slits,
The distance of the 5th bright fringe from the central bright fringe will be given by
B) 8.1 cm
The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:
To find the angle corresponding to the 8th dark fringe, we substitute n=8:
And the distance of the 8th dark fringe from the central bright fringe will be given by
What are the options u can't say that and expect me to know wat u talkin bout
Heat your answer is Heat.
Hoped I helped.
