Answer:
1.98 rad/s
Explanation:
Given that a spring stretches at a distance 0.1 m when a mass of 0.5 kg is attached to it.
The angular velocity
If it is stretched 0.2 m from the equilibrium length; we have:
The maximum velocity v = 0.2 × 9.9
v = 1.98 rad/s
Therefore; the maximum velocity of the mass = 1.98 rad/s
Answer:
Explanation:
It is a case of oscillation by simple pendulum . Expression for simple pendulum is given as follows
T =
where T is time period , l is length of pendulum and g is acceleration due to gravity .
, f is frequency of oscillation
For the given case
subsequently length becomes half so
dividing
f =
frequency of oscillation becomes √2 times.
Answer
given,
Mass of the solid sphere = 1800 Kg
radius of the sphere,R = 5 m
mass of the small sphere, m = 2.30 Kg
when the Point is outside the sphere the Force between them is equal to
when r>R
When Point is inside the Sphere
when r<R
where r is the distance where the point mass is placed form the center
Now Force calculation
a) r = 5.05 m
[/tex]
F = 1.082 x 10⁻⁸ N
b) r = 2.65 m
