To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

Where,
F = Force
r = Radius
Replacing we have that,



The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore


Finally, angular acceleration is a result of the expression of torque by inertia, therefore



PART B)
The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians
, therefore



Answer:
3.88 * 10^(-15) J
Explanation:
We know that the Potential energy of the electron at the beginning of its motion is equal to the Kinetic energy at the end of its motion, when it reaches the plates.
First, we get the potential and potential energy:
Electric potential = E * r
E = electric field
r = distance between plates
Potential = 2.2 * 10^6 * 0.011
= 2.42 * 10^4 V
The relationship between electric potential and potential energy is:
P. E. = q*V
q = charge of electron = 1.602 * 10^(-19) C
P. E. = 2.42 * 10^4 * 1.602 * 10^(-19)
P. E. = 3.88 * 10^(-15) J
Answer:
f this bs
Explanation:
no i will not answer this
Answer:
The crate's coefficient of kinetic friction on the floor is 0.23.
Explanation:
Given that,
Mass of the crate, m = 300 kg
One worker pushes forward on the crate with a force of 390 N while the other pulls in the same direction with a force of 320 N using a rope connected to the crate.
The crate slides with a constant speed. It means that the net force acting on it is 0. Net force acting on it is given by :

So, the crate's coefficient of kinetic friction on the floor is 0.23.