Answer:
Gravity
Explanation:
Gravity is a force that pulls the surface of the earth and keeps the planets in orbits around the sun.
Answer:
Explanation:
Given
Diameter of Pulley=10.4 cm
mass of Pulley(m)=2.3 kg
mass of book![(m_0)=1.7 kg](https://tex.z-dn.net/?f=%28m_0%29%3D1.7%20kg)
height(h)=1 m
time taken=0.64 s
![h=ut+frac{at^2}{2}](https://tex.z-dn.net/?f=h%3Dut%2Bfrac%7Bat%5E2%7D%7B2%7D)
![1=0+\frac{a(0.64)^2}{2}](https://tex.z-dn.net/?f=1%3D0%2B%5Cfrac%7Ba%280.64%29%5E2%7D%7B2%7D)
![a=4.88 m/s^2and [tex]a=\alpha r](https://tex.z-dn.net/?f=a%3D4.88%20m%2Fs%5E2%3C%2Fp%3E%3Cp%3Eand%20%5Btex%5Da%3D%5Calpha%20r)
where
is angular acceleration of pulley
![4.88=\alpha \times 5.2\times 10^{-2}](https://tex.z-dn.net/?f=4.88%3D%5Calpha%20%5Ctimes%205.2%5Ctimes%2010%5E%7B-2%7D)
![\alpha =93.84 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D93.84%20rad%2Fs%5E2)
And Tension in Rope
![T=m(g-a)](https://tex.z-dn.net/?f=T%3Dm%28g-a%29)
![T=1.7\times (9.8-4.88)](https://tex.z-dn.net/?f=T%3D1.7%5Ctimes%20%289.8-4.88%29)
T=8.364 N
and Tension will provide Torque
![T\times r=I\cdot \alpha](https://tex.z-dn.net/?f=T%5Ctimes%20r%3DI%5Ccdot%20%5Calpha%20)
![8.364\times 5.2\times 10^{-2}=I\times 93.84](https://tex.z-dn.net/?f=8.364%5Ctimes%205.2%5Ctimes%2010%5E%7B-2%7D%3DI%5Ctimes%2093.84)
![I=0.463\times 10^{-2} kg-m^2](https://tex.z-dn.net/?f=I%3D0.463%5Ctimes%2010%5E%7B-2%7D%20kg-m%5E2)
![I_{original}=\frac{mr^2}{2}=0.31\times 10^{-2}kg-m^2](https://tex.z-dn.net/?f=I_%7Boriginal%7D%3D%5Cfrac%7Bmr%5E2%7D%7B2%7D%3D0.31%5Ctimes%2010%5E%7B-2%7Dkg-m%5E2)
Thus mass is uniformly distributed or some more towards periphery of Pulley
Answer:
The average magnetic flux through each turn of the inner solenoid is ![11.486\times10^{-8}\ Wb](https://tex.z-dn.net/?f=11.486%5Ctimes10%5E%7B-8%7D%5C%20Wb)
Explanation:
Given that,
Number of turns = 22 turns
Number of turns another coil = 330 turns
Length of solenoid = 21.0 cm
Diameter = 2.30 cm
Current in inner solenoid = 0.140 A
Rate = 1800 A/s
Suppose For this time, calculate the average magnetic flux through each turn of the inner solenoid
We need to calculate the magnetic flux
Using formula of magnetic flux
![\phi=BA](https://tex.z-dn.net/?f=%5Cphi%3DBA)
![\phi=\dfrac{\mu_{0}N_{2}I}{l}\times\pi r^2](https://tex.z-dn.net/?f=%5Cphi%3D%5Cdfrac%7B%5Cmu_%7B0%7DN_%7B2%7DI%7D%7Bl%7D%5Ctimes%5Cpi%20r%5E2)
Put the value into the formula
![\phi=\dfrac{4\pi\times10^{-7}\times330\times0.140}{21.0\times10^{-2}}\times\pi\times(\dfrac{2.30\times10^{-2}}{2})^2](https://tex.z-dn.net/?f=%5Cphi%3D%5Cdfrac%7B4%5Cpi%5Ctimes10%5E%7B-7%7D%5Ctimes330%5Ctimes0.140%7D%7B21.0%5Ctimes10%5E%7B-2%7D%7D%5Ctimes%5Cpi%5Ctimes%28%5Cdfrac%7B2.30%5Ctimes10%5E%7B-2%7D%7D%7B2%7D%29%5E2)
![\phi=11.486\times10^{-8}\ Wb](https://tex.z-dn.net/?f=%5Cphi%3D11.486%5Ctimes10%5E%7B-8%7D%5C%20Wb)
Hence, The average magnetic flux through each turn of the inner solenoid is ![11.486\times10^{-8}\ Wb](https://tex.z-dn.net/?f=11.486%5Ctimes10%5E%7B-8%7D%5C%20Wb)
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