<h2>Answer:</h2>
<u>Acceleration is </u><u>-10.57 rad/s² </u>
<u>Time is </u><u>6 seconds</u>
<h2>Explanation:</h2><h3>a) </h3>
u=900rpm= 94.248 rad/s
v =300rpm= 31.416 rad/s
s=60 revolutions= 377 rad
v²= u²+ 2as
31.416² = 94.248²+ 2 * a * 377
a = v²-u² / 2s
a= -10.57 rad/s²
<h3>b) </h3>
Using 1st equation of motion
v-u/a = t
Putting the values
t = (31.4 - 94.2)/-10.57
t = 6 seconds
Answer:
Explanation:
Let mass of cylinder be M
Moment of inertia of cylinder
= 1/2 M R² r is radius of cylinder
If radius of equivalent hoop be k
Mk² = 1/2 x MR²
k = R / √2
1.2 / 1.414
Radius of gyration = 0.848 m
b )
moment of inertia of spherical shell
= 2 / 3 M R²
Moment of inertia of equivalent hoop
Mk²
So
Mk² = 2 / 3 M R²
k = √2/3 x R
= .816 X 1.2
Radius of gyration = .98 m
c )
Moment of inertia of solid sphere
= 2/5 M R²
Moment of inertia of equivalent hoop
= Mk²
Mk² = 2/5 M R²
k √ 2/5 R
Radius of gyration = .63 R
Answer:
Explanation:
Given that,
We have two capacitors connected in series
C1=2.0-μF
C2=4.0-μF
Then the equivalent of their series connection
1/Ceq = ½ + ¼
1/Ceq= (2+1)/4
1/Ceq=¾
Taking the reciprocal
Ceq= 4/3 μF
The capacitors are connected to a battery of 1kv
V=1000Volts
We know that,
Q=CV
Where Q is charge
C is capacitance and
V is voltage
Then, Q=4/3 ×1000
Q=4000/3 -μC
Since the capacitors are in series, then the charge pass through them, so each charge on the capacitors are 4000/3 μF
After the capacitor has been charge, the capacitor are disconnect and reconnected in parallel to each other,
For parallel connection, they have the same voltage but different charges.
When connected in parallel, there is a charge redistribution,
And the total charge will be 2•4000/3=8000/3 -μF
Then, Q1 +Q2= 8000/3 μF
Now the charge on each capacitor will be, let them have a common voltage V
Q=CV
Then, Q1=C1V
Q1= 2×V=2V
Q2= 4×V=4V
Then, Q1+Q2=8000/3
4V+2V=8000/3
6V=8000/3
V=8000/(3×6)
V=4000/9
V=444.44Volts
Now, Q1=2V
Q1=2×4000/9
Q1=8000/9 μF
Also, Q2=4V
Q2=4×4000/9
Q2=16000/9 μF
There is not enough information to answer the question
