Creatine phosphate can supply the energy needs of a working muscle at a very high rate, but only for about 8–10 seconds.
The complex, highly technical formula for capacitors is
<em>Q = C V</em>
Charge = (capacitance) (voltage)
Charge = (3 F) (24 V)
<em>Charge = 72 Coulombs</em>
The positive plate of the capacitor is missing 72 coulombs worth of electrons. They were sucked into positive terminal of the battery stack.
The negative plate of the capacitor has 72 coulombs worth of extra electrons. They came from the negative terminal of the battery stack.
You should be aware that this is a humongous amount of charge ! An average <u><em>lightning bolt</em></u>, where electrons flow between a cloud and the ground for a short time, is estimated to transfer around <u><em>15 coulombs</em></u> of charge !
The scenario in the question involves a "supercapacitor". 3 F is is no ordinary component ... One distributor I checked lists one of these that's able to stand 24 volts on it, but that product costs $35 apiece, you have to order at least 100 of them at a time, and they take 2 weeks to get.
Also, IF you can charge this animal to 24 volts, it will hold 864J of energy. You'd probably have a hard time accomplishing this task with a bag of leftover AA batteries.
Answer:
The child will take 5.952 seconds to travel from the top of the hill to the bottom.
Explanation:
Given that the child accelerates uniformly and that both initial (
) and final speeds (
), measured in meters per second, and acceleration (
), measured in meters per square second, are known, we proceed to use the following kinematic equation to determine the time taken to travel from the top of the hill to the bottom (
), measured in seconds, is:
(1)
If we know that 
, 
and 
, then the time taken is:
The child will take 5.952 seconds to travel from the top of the hill to the bottom.
Answer:
t=7.33 s
Explanation:
According to Newton's second law:
because we don't want the box to slide, the acceleration has to be zero.
we know that:
Now having the acceleration, we can use the following formula.
Answer:
Explanation:
Inductance L = 1.4 x 10⁻³ H
Capacitance C = 1 x 10⁻⁶ F
a )
current I = 14 .0 t
dI / dt = 14
voltage across inductor
= L dI / dt
= 1.4 x 10⁻³ x 14
= 19.6 x 10⁻³ V
= 19.6 mV
It does not depend upon time because it is constant at 19.6 mV.
b )
Voltage across capacitor
V = ∫ dq / C
= 1 / C ∫ I dt
= 1 / C ∫ 14 t dt
1 / C x 14 t² / 2
= 7 t² / C
= 7 t² / 1 x 10⁻⁶
c ) Let after time t energy stored in capacitor becomes equal the energy stored in capacitance
energy stored in inductor
= 1/2 L I²
energy stored in capacitor
= 1/2 CV²
After time t
1/2 L I² = 1/2 CV²
L I² = CV²
L x ( 14 t )² = C x ( 7 t² / C )²
L x 196 t² = 49 t⁴ / C
t² = CL x 196 / 49
t = 74.8 μ s
After 74.8 μ s energy stored in capacitor exceeds that of inductor.
