The shades are very different
The weight of the meterstick is:
![W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N](https://tex.z-dn.net/?f=W%3Dmg%3D0.20%20kg%20%5Ccdot%209.81%20m%2Fs%5E2%20%3D%201.97%20N)
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
![d_1 = 0.50 m - 0.40 m=0.10 m](https://tex.z-dn.net/?f=d_1%20%3D%200.50%20m%20-%200.40%20m%3D0.10%20m)
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
![M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm](https://tex.z-dn.net/?f=M_w%20%3D%20W%20d_1%20%3D%20%281.97%20N%29%280.10%20m%29%3D0.20%20Nm)
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
![(mg) d_2 = 0.20 Nm](https://tex.z-dn.net/?f=%28mg%29%20d_2%20%3D%200.20%20Nm)
from which we find the value of d2:
![d_2 = \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m](https://tex.z-dn.net/?f=d_2%20%3D%20%20%5Cfrac%7B0.20%20Nm%7D%7Bmg%7D%3D%20%5Cfrac%7B0.20%20Nm%7D%7B%280.5%20kg%29%289.81%20m%2Fs%5E2%29%7D%3D0.04%20m%20%20)
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Answer:
t = 5.56 ms
Explanation:
Given:-
- The current carried in, Iin = 1.000002 C
- The current carried out, Iout = 1.00000 C
- The radius of sphere, r = 10 cm
Find:-
How long would it take for the sphere to increase in potential by 1000 V?
Solution:-
- The net charge held by the isolated conducting sphere after (t) seconds would be:
qnet = (Iin - Iout)*t
qnet = t*(1.000002 - 1.00000) = 0.000002*t
- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:
V = k*qnet / r
Where, k = 8.99*10^9 ..... Coulomb's constant
qnet = V*r / k
t = 1000*0.1 / (8.99*10^9 * 0.000002)
t = 5.56 ms