Answer:
962 rpm.
Explanation:
given,
angular acceleration = 190 rad/s²
initial angular speed = 0 rad/s
final angular speed = 7200 rpm
=
=
we need to calculate the revolution of disk after 10 s.
time taken to reach the final angular velocity
using equation of angular motion


t = 4 s
rotation of wheel in 4 s



θ = 1520 rad


now, revolution of the disk in next 6 s
angular velocity is constant


θ_f = 6044 rad
θ_f = 
revolution of the computer hard disk
θ_f = 962 rpm.
total revolution of the computer disk after 10 s is equal to 962 rpm.
Answer:
119.42 x 10⁻³ m
Explanation:
The distance of first dark fringe from central bright spot will be equal to the width of fringe .
Fringe width = λL /w
where λ is wavelength of light, w is slit width , L is distance of screen .
So required distance
= 
= 119.42 x 10⁻³ m
Answer:
Explanation:
Angular velocity of satellite
= 2π x .01
= .02 π rad /s
Initial angular momentum
Moment of inertia x angular velocity
= 2000 x .02 π
= 125.6 unit
Linear impulse produced by each thruster
= 15 N.s
Angular impulse
= 15 x 1.5 = 22.5 unit
Total angular impulse in 30 pulses
= 22.5 x 2 x 30
1350
This angular impulse will add total angular momentum of
1350 unit
So total angular momentum after 30 pulses
= 1350 + 22.5
= 1372.5 unit
So final angular velocity
= final angular momentum / moment of inertia
= 1372.5 / 2000
= 0 .686 rad /s
The correct answer to the question is B) 5,700 years.
EXPLANATION:
Before going to answer this question, first we have to understand half life period of a radio active substance.
The half life period
of a radio active substance is defined as the time in which half of the radio active specimen has undergone decay.
As per the question, the initial concentration of the radioactive carbon-14 is given as 10,000.
Hence, initial concentration
.
After 5,700 years, the remaining radio active specimen is 5000. It means that half of the radio active specimen has undergone decay in 5,700 years.
Hence, half life period of the radio active carbon-14 specimen is 5,700 years.