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
=> 
=>
Efficiency = Work Output / Work Input
92% = Work Output / 100
0.92 = Work Output / 100
Work Output = 0.92 * 100
Work Output = 92 joules.
Erosion, weathering, mechanical changes, chemical changes.
Really, any interaction can change the composition of a rock whether it be done by man or through nature.
The angle of the ladder inclined with respect to the horizontal after being moved a distance of 0.82 m closer to the building is 53.84°
cos θ = Adjacent side / Hypotenuse
θ
= 47°
Hypotenuse = Length of ladder = 8.5 m
cos 47° = Adjacent side / 8.5
Adjacent side = Initial distance of base of ladder from the building = 5.8 m
Adjacent side 2 = Final distance of base of ladder from the building
Adjacent side 2 = 5.8 - 0.82 = 4.98 m
cos θ
= Adjacent side 2 / Hypotenuse
cos θ = 4.98 / 8.5 = 0.59
θ = 
( 0.59 )
θ = 53.84°
The formula used above is one of trigonometric ratios. Trigonometric ratios can used only in a right angled triangle where one of the angles in at 90 degrees and the other two angles are less than 90 degrees.
Therefore, the angle of the ladder inclined with respect to the horizontal after being moved is 53.84°
To know more about trigonometric ratios
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