Explanation:
a) From Coulombs law,
F = (kq1q2)/r²...............1
F = 10 N
r = 10m
k = 9 × 10^9 m/F
q1 = q2
from equation 1, make q the subject
q = (Fr²)/2k
q = (10×10²)/(2×9×10^9)
q = 1000/1.8×10^10
q = 5.556×10^-8 C
b) K = k/ko
k = Kko = 10times permittivity of free-space
Goodluck
Explanation:
We need to calculate the centripetal force:
Fc = W + F
With Fc being the centripetal force, W the weight of the boy, F the centrifugal force (apparent).
We know that we can calculate the apparent centrifugal force thank to the formula:
F = (m·v²)/r = 204N
So we can write:
Fc = W + F = 445N + 204N = 649N
Answer:
695800 N/m^2 or Pa
Explanation:
Height of the water from the ground H = 71 m
Acceleration due to gravity g =9.8 m/s^2
density of water ρ= 1000 kg/m^3
The minimum output gauge pressure to make water reach height H
P= ρgH
= 1000×9.8×71= 695800 N/m^2 or Pa
I think:
In motion- 40
Not moving- 20
