Answer:
Greater than
Explanation:
Here, angular momentum is conserved.
![l_1\omega_1 =l_2\omega_2](https://tex.z-dn.net/?f=l_1%5Comega_1%20%3Dl_2%5Comega_2)
When the cloud shrinks under the right conditions, a star may be formed.
Thus, Diameter of clouds are much higher than a star.
Moment of inertia of cloud is greater than the star's inertial.
so, angular velocity of the star would be greater than angular velocity of the rotating gas.
Answer:
1➡️ this is the method of decomposition
2➡️ H2 and O2
3➡️ b
sorry if I am wrong
Answer:
Form of energy: Example
1. Light energy: Electromagnetic radiation
2. Nuclear energy: Nuclear fission
3. Chemical energy: Energy stored in plant matter
4. Electrical energy: Lightning
5. Thermal energy: A hot surface
6. Sound energy: A tuning fork
7. Solar energy: Energy from the Sun
8. Mechanical energy: A moving vehicle
Answer:
A) ![\omega_f=17.503\ rad.s^{-1}](https://tex.z-dn.net/?f=%5Comega_f%3D17.503%5C%20rad.s%5E%7B-1%7D)
B) ![t=55.6822\ s](https://tex.z-dn.net/?f=t%3D55.6822%5C%20s)
C) ![\theta=1312\ rad](https://tex.z-dn.net/?f=%5Ctheta%3D1312%5C%20rad)
Explanation:
Given:
- mass of flywheel,
![m=40\ kg](https://tex.z-dn.net/?f=m%3D40%5C%20kg)
- diameter of flywheel,
![d=0.72\ m](https://tex.z-dn.net/?f=d%3D0.72%5C%20m)
- rotational speed of flywheel,
![N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}](https://tex.z-dn.net/?f=N_i%3D450%5C%20rpm%20%5CRightarrow%20%5Comega_i%3D%5Cfrac%7B450%5Ctimes%202%5Cpi%7D%7B60%7D%20%3D15%5Cpi%5C%20rad.s%5E%7B-1%7D)
- duration for which the power is off,
![t_0=35\ s](https://tex.z-dn.net/?f=t_0%3D35%5C%20s)
- no. of revolutions made during the power is off,
![\theta=180\times 2\pi=360\pi\ rad](https://tex.z-dn.net/?f=%5Ctheta%3D180%5Ctimes%202%5Cpi%3D360%5Cpi%5C%20rad)
<u>Using equation of motion:</u>
![\theta=\omega_i.t+\frac{1}{2} \alpha.t^2](https://tex.z-dn.net/?f=%5Ctheta%3D%5Comega_i.t%2B%5Cfrac%7B1%7D%7B2%7D%20%5Calpha.t%5E2)
![360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2](https://tex.z-dn.net/?f=360%5Cpi%3D15%5Cpi%5Ctimes%2035%2B%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Calpha%5Ctimes35%5E2)
![\alpha=-0.8463\ rad.s^{-2}](https://tex.z-dn.net/?f=%5Calpha%3D-0.8463%5C%20rad.s%5E%7B-2%7D)
Negative sign denotes deceleration.
A)
Now using the equation:
![\omega_f=\omega_i+\alpha.t](https://tex.z-dn.net/?f=%5Comega_f%3D%5Comega_i%2B%5Calpha.t)
![\omega_f=15\pi-0.8463\times 35](https://tex.z-dn.net/?f=%5Comega_f%3D15%5Cpi-0.8463%5Ctimes%2035)
is the angular velocity of the flywheel when the power comes back.
B)
Here:
![\omega_f=0\ rad.s^{-1}](https://tex.z-dn.net/?f=%5Comega_f%3D0%5C%20rad.s%5E%7B-1%7D)
Now using the equation:
![\omega_f=\omega_i+\alpha.t](https://tex.z-dn.net/?f=%5Comega_f%3D%5Comega_i%2B%5Calpha.t)
![0=15\pi-0.8463\times t](https://tex.z-dn.net/?f=0%3D15%5Cpi-0.8463%5Ctimes%20t)
is the time after which the flywheel stops.
C)
Using the equation of motion:
![\theta=\omega_i.t+\frac{1}{2} \alpha.t^2](https://tex.z-dn.net/?f=%5Ctheta%3D%5Comega_i.t%2B%5Cfrac%7B1%7D%7B2%7D%20%5Calpha.t%5E2)
![\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2](https://tex.z-dn.net/?f=%5Ctheta%3D15%5Cpi%5Ctimes%2055.68225-0.5%5Ctimes%200.8463%5Ctimes%2055.68225%5E2)
revolutions are made before stopping.