In each of your hands you hold a smartphone one has a mass of 0.130kg and the other has a mass of 0.200kg. if both phones create
1 answer:
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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:
The resultant vector is 1 m/s
Explanation:
The resultant vector is 1 m/s west based on triangle law of vector addition, when two sides of a triangle is represented by two vectors, the resultant vector is the third side of the triangle.
Answer:
hello your question is incomplete attached below is the missing part
answer : short period oscillations frequency = 0.063 rad / sec
phugoid oscillations natural frequency ( ) = 4.27 rad/sec
Explanation:
first we have to state the general form of the equation
=
where :
comparing the general form with the given equation
= 18.2329
hence the short period oscillation frequency ( ) = 0.063 rad/sec
phugoid oscillations natural frequency ( ) = 4.27 rad/sec
