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:
Total length of spring 0.647 m
Explanation:
We have given mass of the person m = 150 kg
Acceleration due to gravity
Spring constant k = 10000 N/m
Nominal length of spring = 0.50
According to hook's law
x = 0.147 m
So total length of spring = 0.50+0.147 = 0.647 m
Answer:a goal is scored by throwing the ball into the goal post and is confirmed by deciding that’s if the ball has hit the tape mark or not
Explanation:
To solve this problem we will use the trigonometric concepts to find the distance h, which will allow us to find the speed of Jeff and that will finally be the variable that will indicate the total tension, since it is the variable of the centrifugal Force given in the vine at the lowest poing of the swing.
From the image:
When Jeff reaches his lowest point his potential energy is converted to kinetic energy
Tension in the string at the lowest point is sum of weight of Jeff and the his centripetal force
Therefore the tension in the vine at the lowest point of the swing is 842.49N