Answer:

0.3619sec
Explanation:
Given that
Mass,m=148 g
Length,L=13 cm
Velocity,u'(0)=10 cm/s
We have to find the position u of the mass at any time t
We know that

Where 

u(0)=0
Substitute the value

Substitute u'(0)=10


Substitute the values

Period =T = 2π/8.68
After half period
π/8.68 it returns to equilibruim
π/8.68 = 0.3619sec
The kinetic energy is 945 joules.
Kinetic energy is the energy that an object has as a result of motion. It is defined as the effort required to accelerate a mass-determined body from rest to the indicated velocity.
The speed of an object or particle, which is a scalar quantity, is the size of the change in its location over time or the size of the change in its position per unit of time.
The mass of the volleyball is 2.1 kg.
The speed of the ball when the ball leaves the hand is 30 m/s.
m = 2.1 kg
v = 30 m/s
The kinetic energy of an object is given as:
KE = (1/2 ) × m × v²
KE = (1 / 2) × 2.1 kg × ( 30 m/s)²
KE = (1 / 2) × 2.1 kg × 30 m/s × 30 m/s
KE = 2.1 kg × 15 m/s × 30 m/s
KE = 945 J
Learn more about kinetic energy here:
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Answer:
1.045 m from 120 kg
Explanation:
m1 = 120 kg
m2 = 420 kg
m = 51 kg
d = 3 m
Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.
By use of the gravitational force
Force on m due to m1 is equal to the force on m due to m2.



3 - y = 1.87 y
3 = 2.87 y
y = 1.045 m
Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.
Answer
given,
diameter of the pipe is = (14 ft)4.27 m
minimum speed of the skater must have at very top = ?
At the topmost point of the pipe the normal force will be equal to zero.
F = mg
centripetal force acting on the skateboard

equating both the force equation


r = d/2 = 14/ 2 = 7 ft
or
r = 4.27/2 = 2.135 m
g = 32 ft/s² or g = 9.8 m/s²

v = 14.96 ft/s
or

v = 4.57 m/s
The position of the first ball is

while the position of the second ball, thrown with initial velocity
, is

The time it takes for the first ball to reach the halfway point satisfies



We want the second ball to reach the same height at the same time, so that



