Answer:
a) t=24s
b) number of oscillations= 11
Explanation:
In case of a damped simple harmonic oscillator the equation of motion is
m(d²x/dt²)+b(dx/dt)+kx=0
Therefore on solving the above differential equation we get,
x(t)=A₀
where A(t)=A₀
A₀ is the amplitude at t=0 and
is the angular frequency of damped SHM, which is given by,

Now coming to the problem,
Given: m=1.2 kg
k=9.8 N/m
b=210 g/s= 0.21 kg/s
A₀=13 cm
a) A(t)=A₀/8
⇒A₀
=A₀/8
⇒
applying logarithm on both sides
⇒
⇒
substituting the values

b) 

, where
is time period of damped SHM
⇒
let
be number of oscillations made
then, 
⇒
Answer:
<em>The force required is 3,104 N</em>
Explanation:
<u>Force</u>
According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:
F = ma
Where a is the acceleration of the object.
On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

Where:
vf is the final speed
vo is the initial speed
a is the acceleration
t is the time
Solving for a:

We are given the initial speed as vo=20.4 m/s, the final speed as vf=0 (at rest), and the time taken to stop the car as t=7.4 s. The acceleration is:


The acceleration is negative because the car is braking (losing speed). Now compute the force exerted on the car of mass m=1,126 kg:

F= 3,104 N
The force required is 3,104 N
Answer:
an armature a permanent magnet brushes slip rings
Explanation:
Answer:
Would love to be your friend m'lady.
Explanation: