Answer:
It will cause kinetic energy to increase.
Explanation:
Given that Speed and Motion you went from the starting line to the finish line at different rates.
If you repeated the activity while carrying weights but keeping your times the same, the weight carried will add up to the mass of the body.
And since Kinetic energy K.E = 1/2mv^2
Increase in the mass of the body will definitely make the kinetic energy of the body to increase.
Since the time is the same, that means the speed V is the same.
Weight W = mg
m = W/g
The new kinetic energy will be:
K.E = 1/2(M + m)v^2
This means that there will be increase in kinetic energy.
Answer:
The new force is 1/4 of the previous force.
Explanation:
Given
---- ![r_1](https://tex.z-dn.net/?f=r_1)
--- ![r_2](https://tex.z-dn.net/?f=r_2)
Required
Determine the new force
Let the two particles be q1 and q2.
The initial force F1 is:
--- Coulomb's law
Substitute 2 for r1
![F_1 = \frac{kq_1q_2}{2^2}](https://tex.z-dn.net/?f=F_1%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7B2%5E2%7D)
![F_1 = \frac{kq_1q_2}{4}](https://tex.z-dn.net/?f=F_1%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7B4%7D)
The new force (F2) is
![F_2 = \frac{kq_1q_2}{r_2^2}](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7Br_2%5E2%7D)
Substitute 4 for r2
![F_2 = \frac{kq_1q_2}{4^2}](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7B4%5E2%7D)
![F_2 = \frac{kq_1q_2}{4*4}](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7B4%2A4%7D)
![F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7B1%7D%7B4%7D%2A%5Cfrac%7Bkq_1q_2%7D%7B4%7D)
Substitute ![F_1 = \frac{kq_1q_2}{4}](https://tex.z-dn.net/?f=F_1%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7B4%7D)
![F_2 = \frac{1}{4}*F_1](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7B1%7D%7B4%7D%2AF_1)
![F_2 = \frac{F_1}{4}](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7BF_1%7D%7B4%7D)
The new force is 1/4 of the previous force.
The expression of the electric flux is
![\Phi = \frac{Q}{\epsilon_0}](https://tex.z-dn.net/?f=%5CPhi%20%3D%20%5Cfrac%7BQ%7D%7B%5Cepsilon_0%7D)
Here,
Q = Total charge enclosed in the closed surface
= Permittivity due to free space
Rearranging to find the charge,
![Q = \epsilon_0 \Phi](https://tex.z-dn.net/?f=Q%20%3D%20%5Cepsilon_0%20%5CPhi)
Replacing with our values we have finally
![Q = (8.85*10^{-12}F\cdot m^{-1})(1.84*10^3 N\cdot m^2/C)](https://tex.z-dn.net/?f=Q%20%3D%20%288.85%2A10%5E%7B-12%7DF%5Ccdot%20m%5E%7B-1%7D%29%281.84%2A10%5E3%20N%5Ccdot%20m%5E2%2FC%29)
![(\frac{10^9nC}{1C})](https://tex.z-dn.net/?f=%28%5Cfrac%7B10%5E9nC%7D%7B1C%7D%29)
![Q = 0.1684nC](https://tex.z-dn.net/?f=Q%20%3D%200.1684nC)
The charge enclosed by the box is 0.1684nC
The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.
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