Kinetic energy = mass time squared speed divided by 2
W=mv^2/2 = 50*10*10/2 = 2500 J
A right turn would be the answer probably.
Explanation:
Below is an attachment containing the solution.
Answer:
Explanation:
Given that,
First Capacitor is 10 µF
C_1 = 10 µF
Potential difference is
V_1 = 10 V.
The charge on the plate is
q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC
q_1 = 100 µC
A second capacitor is 5 µF
C_2 = 5 µF
Potential difference is
V_2 = 5V.
Then, the charge on the capacitor 2 is.
q_2 = C_2 × V_2
q_2 = 5µF × 5 = 25 µC
Then, the average capacitance is
q = (q_1 + q_2) / 2
q = (25 + 100) / 2
q = 62.5µC
B. The two capacitor are connected together, then the equivalent capacitance is
Ceq = C_1 + C_2.
Ceq = 10 µF + 5 µF.
Ceq = 15 µF.
The average voltage is
V = (V_1 + V_2) / 2
V = (10 + 5)/2
V = 15 / 2 = 7.5V
Energy dissipated is
U = ½Ceq•V²
U = ½ × 15 × 10^-6 × 7.5²
U = 4.22 × 10^-4 J
U = 422 × 10^-6
U = 422 µJ
Answer:
W = 1080.914 J
Explanation:
f(x) = 1100xe⁻ˣ
Work done by a variable force moving through a particular distance
W = ∫ f(x) dx (with the integral evaluated between the interval that the force moves through)
W = ∫⁶₀ 1100xe⁻ˣ dx
W = 1100 ∫⁶₀ xe⁻ˣ dx
But the integral can only be evaluated using integration by parts.
∫ xe⁻ˣ dx
∫ vdu = uv - ∫udv
v = x
(dv/dx) = 1
dv = dx
du = e⁻ˣ dx
∫ du = ∫ e⁻ˣ dx
u = -e⁻ˣ
∫ vdu = uv - ∫udv
∫ xe⁻ˣ dx = (-e⁻ˣ)(x) - ∫ (-e⁻ˣ)(dx)
= -xe⁻ˣ - e⁻ˣ = -e⁻ˣ (x + 1)
∫ xe⁻ˣ dx = -e⁻ˣ (x + 1) + C (where c = constant of integration)
W = 1100 ∫⁶₀ xe⁻ˣ dx
W = 1100 [-e⁻ˣ (x + 1)]⁶₀
W = 1100 [-e⁻⁶ (6 + 1)] - [-e⁰ (0 + 1)]
W = 1100 [-0.0173512652 + 1]
W = 1100 × (0.9826487348)
W = 1080.914 J
Hope this Helps!!!
