Answer:
a) 20.81 J
b) 8.29 J
Explanation:
V = iR + L di/dt
where
i = a(1-e^-kt)
for large t
i = V/R
i = 24 / 9.4
i = 2.55 A
so
i = 2.55(1-e^-kt)
di/dt = 2.55 k e^-kt
24 = 24-24e^-kt + 6.4(2.55)k e^-kt
24 = 6.4(2.55) k
k = 24 / (6.4 * 2.55)
k = 24 / 16.32
k = 1.47 = R/L
so
i = 2.55(1-e^-(Rt/L))
current is maximum at great t
i max = 2.55 - 0
energy = (1/2) L i^2
E = (1/2)(6.4)2.55^2
E = 20.81 Joules
one time constant T = L/R and e^-(Rt/L) = 1/e = .368
i = 2.55 (1 - 0.368)
i = 2.55 * 0.632
i = 1.61 amps
energy = (1/2)(6.4)1.61^2
E = 8.29 Joules
Answer:
a. I = -3mv₀/2 b. F = -3mv₀/2t
Explanation:
a. Use the impulse-momentum to write an equation for the system which is the cart only.
We know Impulse , I = Δp where Δp = change in momentum.
Now, Δp = m(v - u) where m = mass of cart, u = initial velocity of cart = v₀ and v = rebound velocity of cart = -v₀/2 (negative since it moves in the opposite direction to u)
So, I = Δp = m(v - u)
Substituting the values of the variables into the equation, we have
I = m(-v₀/2 - v₀) = -3mv₀/2
So, I = -3mv₀/2
b. If the time during which the bumper exerts a force on the cart is t, write an expression for the force F exerted on the cart in terms of the given variables.
We know impulse I = Ft where F = force exerted on the cart and t = time force acts
Also, I = -3mv₀/2
So, Ft = -3mv₀/2
F = -3mv₀/2t
Answer:
F = 0.408 N
Explanation:
It is given that,
Weight of a wooden block, W = 8 N
Weight, W = mg
m is mass of wooden block
Acceleration of the block, a = 0.5 m/s²
Force, F = ma
So, the magnitude of force applied to the wooden block is 0.408 N.
The force produced on a particle of charge q by an electric field of intensity E is
in our problem, the force is F=5.0 N while the charge is q=2.0 C, so we can find the intensity of the electric field:
The relationship between electric field intensity and potential difference
between two points A and B is
where d is the distance between the two points. By using d=6.0 m, we find
where the negative sign means that the initial point, VA, is at higher potential than the final point VB.
Answer:
This is due to the gap in between the switch. So the applied energy is not converted to current even though the resistance is still there. Hence voltage will be there, where you applied it
Explanation: