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Yuri [45]
3 years ago
13

An inductor is connected in series to a fully charged capacitor. Which of the following statements are true? Check all that appl

y.- As the capacitor is charging, the current is increasing.- The stored electric field energy can be greater than the stored magnetic field energy.- As the capacitor is discharging, the current is increasing.- The stored electric field energy can be less than the stored magnetic field energy.- The stored electric field energy can be equal to the stored magnetic field energy.
Physics
1 answer:
Luba_88 [7]3 years ago
3 0

Answer:

As the capacitor is discharging, the current is increasing

Explanation:

Lets take

C= Capacitance

L=Inductance

V=Voltage

I= Current

The total energy E given as

E=\dfrac{IL^2}{2}+\dfrac{CV^2}{2}

We know that total energy E is conserved so when electric energy 1/2 CV² decreases then magnetic energy 1/2 IL²  will increases.

It means that when charge on the capacitor decreases then the current will increase.

As the capacitor is discharging, the current is increasing

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Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an
joja [24]

Answer:

Approximately 9.7\; \rm m \cdot s^{-2}.

Explanation:

Assuming that there is no other force on this vehicle, the 16000\; \rm N force from the road would be the only force on this vehicle. The net force would then be equal to this 16000\; \rm N\! force. The size of the net force would be 16000\; \rm N\!\!.

Let m denote the mass of this vehicle and let \Sigma F denote the net force on this vehicle.  

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, a, would be:

\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}.

For this vehicle, \Sigma F = 16000\; \rm N whereas m = 1650\; \rm kg. The acceleration of this vehicle would be:

\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}.

8 0
3 years ago
3. When the procedure is repeated with a third line how will it distinguish whether the location of the center of gravity is acc
Law Incorporation [45]
Well this is question is easy. I mean i’m the one to say the least, The answer is.. SIKE L bozo imagine not knowing the answer
4 0
2 years ago
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drek231 [11]
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4 0
3 years ago
A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (
natita [175]

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

input power = frictional power

frictional power = 2.5 kW

frictional torque x angular velocity = 2.5 kW

frictional torque x .47 = 2.5 kW

frictional torque = 2.5 / .47 kN .m

= 5.32 kN . m

= 5 kN.m

b )

When power is switched off , it will decelerate because of frictional torque .

5 0
3 years ago
How much heat is absorbed by a 71g iron skillet when its temperature rises from 11oC to 29oC?
diamong [38]
The right formula to use for this calculation is the heat capacity formula, 
Heat absorbed, Q = MCT, Where 
M = Mass of the substance = 71
C = Specific heat capacity for iron = 0.450 J/gc
T = Change in temperature = 29 - 11 = 18
Q = 71 * 0.450 *18 = 575.10
The amount of heat absorbed by the iron skillet is 575 J.
5 0
3 years ago
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