Answer:
16.6 ms or 0.0166 s
Explanation:
If Q is the final charge, Q' is the initial charge, C the capacitance ,R is the resistance , t the time taken and τ the time constant,
[tex]Q = Q'( 1- e^{-t\div \tau })
τ = R C = (1.20×10³) (20×10⁻⁶) = 0.024 s
15 = 30 ( 1- e^{-t\div \ 0.024 })
( 1- e^{-t\div \ 0.024 }) = 15 ÷ 30
⇒ - e^{-t\div \ 0.024 }) = 0.5 -1
⇒ e^{-t\div \ 0.024 }) = 0.5
Taking logarithm to the base e on both sides of this equation,
⇒ t = 0.0166 seconds = 16.6 milli seconds
Answer:
Final Length = 30 cm
Explanation:
The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:
F = kΔx
where,
F = Force applied
k = spring constant
Δx = change in length of spring
First, we find the spring constant of the spring. For this purpose, we have the following data:
F = 50 N
Δx = change in length = 25 cm - 20 cm = 5 cm = 0.05 m
Therefore,
50 N = k(0.05 m)
k = 50 N/0.05 m
k = 1000 N/m
Now, we find the change in its length for F = 100 N:
100 N = (1000 N/m)Δx
Δx = (100 N)/(1000 N/m)
Δx = 0.1 m = 10 cm
but,
Δx = Final Length - Initial Length
10 cm = Final Length - 20 cm
Final Length = 10 cm + 20 cm
<u>Final Length = 30 cm</u>
Answer:
Explanation:
There are 3 main forces at work here, gravity, normal and friction. The gravity pulls the car straight down and is what keeps the car on the ground. Normal force is straight up from the points where the car is touching, so since the wheels are the only parts of the car touching the street, this is where all the normal force is. Friction force opposes any and all motion, the car wants to slide down the hill and would slide down the hill if there was no friction, so the friction force is in the opposite direction of the cars intended motion.
