Answer:
u need to make sure that comparison is = to shapes and then find the shapes sizes and add them
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
Answer:
c
Explanation:
a vector quantity has both magnitude and direction
Answer should be <span>t=<span>−<span><span>1<span> or </span></span>t</span></span></span>=<span>5</span>
Answer:
B= 4,000 km
Explanation:
I just took the test on edge :)