Complete question:
A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?
Answer:
The current in the circuit 7 ms later is 0.2499 A
Explanation:
Given;
Ideal inductor, L = 45-mH
Resistor, R = 60-Ω
Ideal voltage supply, V = 15-V
Initial current at t = 0 seconds:
I₀ = V/R
I₀ = 15/60 = 0.25 A
Time constant, is given as:
T = L/R
T = (45 x 10⁻³) / (60)
T = 7.5 x 10⁻⁴ s
Change in current with respect to time, is given as;
Current in the circuit after 7 ms later:
t = 7 ms = 7 x 10⁻³ s
Therefore, the current in the circuit 7 ms later is 0.2499 A
Kinetic energy E = m * v^2
<span>Since the acceleration of both books will be -neglecting air resistance - the same, the kinetic energy will be directly proportional to the mass of the book.</span>
40% the formula is work output/work input*100 so
200/500=.4*100=40
This question involves the conservation of energy. There are two energy in this case, potential energy and kinetic energy. Let's divid the energy into three status.
1. Before dropping, all potential energy
2.dropping, potential energy transformed to kinetic energy
3. before hitting the ground, all Kinetic energy.
Recall the formula for both energy, which are U=mgh, and K=1/2mv^2
Since the energy is conserved in this case ( b/c otherwise it will say in the problem), the amount of energy at the beginning should equal to the energy at the end. Therefore we have, mgh=1/2mv^2
plug the number in and solve for velocity.
2x400x10=1/2 x 2 x v^2
v^2=8000
v=
v=40
