Answer:
The maximum velocity is 1.58 m/s.
Explanation:
A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.
Spring constant, K = 100 N/m
mass, m = 0.1 kg
Amplitude, A = 5 cm = 0.05 m
Let the angular frequency is w.

The maximum velocity is

Answer:
<em>J=36221 Kg.m/s</em>
Explanation:
<u>Impulse-Momentum Theorem</u>
These two magnitudes are related in the following way. Suppose an object is moving at a certain speed
and changes it to
. The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by

The initial and final momentums are, respectively

The change of momentum is

It is numerically equal to the Impulse J


We are given

The impulse the car experiences during that time is

J=-36221 Kg.m/s
The magnitude of J is
J=36221 Kg.m/s
Answer:
The correct one is that the force on B is half of the force on A
Explanation:
Because radius for the inside of the curve is half the radius for the outside and Car A travels on the inside while car B, travels at equal speed on the outside of the curve. Thus force on B will be half on A
As its charge, proton -a positive charged molecule at the center of an atom- is the opposite of the electron -the particle which is orbiting the center of an atom.