A simple pendulum has a bob of mass M. The bob is on a light string of length . The string is fixed at C. At position A, the string is horizontal and the bob is at rest. The bob is released from A and swings to B, where the string is vertical. What is the velocity of the bob when it is at B?

Explanation:

mass of bob = M

string is fixed at C, at position A the string is horizontal and at position B teh string is vertical.

Let the length of the string is L.

At the point C, it has maximum potential energy which is equal to

U = M x g x L   ..... (1)

At the position B, it has maximum kinetic energy and the velocity is v.

K = 1/2 Mv²     ...... (2)

According to the conservation of energy

The potential energy at the position A is equal to the kinetic energy at position B.

M x g x L = 1/2 M x v²

v² = 2 x g x L

the focus of lesson 1 is newton's first law of motion - sometimes referred to as the law of inertia. newton's first law of motion is often stated as. an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

explanation:

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures