Lets say sphere 1 has a charge of 12 + and sphere 2 has a charge of 0 +. After they are touched Sphere 1 becomes 6 + and sphere 2 6 +. So 6 - 12 = a change of 6 -, while 6 - 0 = a change of 6 + Therfore,
Answer: The sign of the charge change / transfered are opposites.
Answer:
a) FE = 0.764FG
b) a = 2.30 m/s^2
Explanation:
a) To compare the gravitational and electric force over the particle you calculate the following ratio:
(1)
FE: electric force
FG: gravitational force
q: charge of the particle = 1.6*10^-19 C
g: gravitational acceleration = 9.8 m/s^2
E: electric field = 103N/C
m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg
You replace the values of all parameters in the equation (1):
![\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764](https://tex.z-dn.net/?f=%5Cfrac%7BF_E%7D%7BF_G%7D%3D%5Cfrac%7B%281.6%2A10%5E%7B-19%7DC%29%28103N%2FC%29%7D%7B%282.2%2A10%5E%7B-18%7Dkg%29%289.8m%2Fs%5E2%29%7D%5C%5C%5C%5C%5Cfrac%7BF_E%7D%7BF_G%7D%3D0.764)
Then, the gravitational force is 0.764 times the electric force on the particle
b)
The acceleration of the particle is obtained by using the second Newton law:
![F_E-F_G=ma\\\\a=\frac{qE-mg}{m}](https://tex.z-dn.net/?f=F_E-F_G%3Dma%5C%5C%5C%5Ca%3D%5Cfrac%7BqE-mg%7D%7Bm%7D)
you replace the values of all variables:
![a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%281.6%2A10%5E%7B-19%7DC%29%28103N%2FC%29-%282.2%2A10%5E%7B-18%7Dkg%29%289.8m%2Fs%5E2%29%7D%7B2.2%2A10%5E%7B-18%7Dkg%7D%5C%5C%5C%5Ca%3D-2.30%5Cfrac%7Bm%7D%7Bs%5E2%7D)
hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.
Answer:
The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².
Explanation:
The centripetal acceleration is given by:
Where:
: is the tangential speed = 9.50 m/s
r: is the distance = 6.00 m
Hence, the centripetal acceleration is:
![a_{c} = \frac{v^{2}}{r} = \frac{(9.50 m/s)^{2}}{6.00 m} = 15.04 m/s^{2}](https://tex.z-dn.net/?f=%20a_%7Bc%7D%20%3D%20%5Cfrac%7Bv%5E%7B2%7D%7D%7Br%7D%20%3D%20%5Cfrac%7B%289.50%20m%2Fs%29%5E%7B2%7D%7D%7B6.00%20m%7D%20%3D%2015.04%20m%2Fs%5E%7B2%7D%20)
Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².
I hope it helps you!