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Goshia [24]
3 years ago
12

Does the charge that flows into the capacitor during the charging go all the way through the capacitor and back to the battery,

or does it get stored somewhere in the capacitor? it goes all the way through the capacitor and back to the battery. it gets stored somewhere in the capacitor.
Physics
2 answers:
Marina CMI [18]3 years ago
8 0
<span>The charge that flows into the capacitor during the charging does not go through the capacitor and back to the battery.   It </span>gets stored somewhere in capacitor, specifically on the plate of the capacitor where the voltage is connected to.  
krok68 [10]3 years ago
4 0

Answer:

it gets stored somewhere in the capacitor.

Explanation:

Capacitor is a charge storage device in which charge is stored on the opposite facing plates

the maximum charge that is stored on the plates of capacitor is given as

Q = CV

here we know that

C = capacitance of the capacitor

V = voltage difference across it

Now when capacitor got its maximum charge then flow of charge through the battery stopped.

So here we can say that correct answer will be

it gets stored somewhere in the capacitor.

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Circuit A in a house has a voltage of 208 V and is limited by a 40.0-A circuit breaker. Circuit B is at 120.0 V and has a 20.0-A
Setler79 [48]

Given Information:  

Voltage of circuit A = Va = 208 Volts

Current of circuit A = Ia = 40 Amps

Voltage of circuit B = Vb = 120 Volts

Current of circuit B = Ib = 20 Amps

Required Information:  

Ratio of power = Pa/Pb = ?

Answer:  

Ratio of power = Pa/Pb = 52/15

Explanation:  

Power can be calculated using Ohm's law

P = VI

Where V is the voltage and I is the current flowing in the circuit.

The power delivered by circuit A is

Pa = Va*Ia

Pa = 208*40

Pa = 8320 Watts

The power delivered by circuit B is

Pb = Vb*Ib

Pb = 120*20

Pb = 2400 Watts

Therefore, the ratio of the maximum power delivered by circuit A to that delivered by circuit B is

Pa/Pb = 8320/2400

Pa/Pb = 52/15

4 0
3 years ago
Gayle runs at a speed of 3.85 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After s
enot [183]

Answer:

Final velocity at the bottom of hill is 15.56 m/s.

Explanation:

The given problem can be divided into four parts:

1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)

From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:  

p_i = p_f  

m_1u_1 + m_2v_2 = (m_1 + m_2)v

v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)}

v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s  

2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.

The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:  

E(i) = E(f)  

KE(i) + PE(i) = KE(f) + PE(f)  

0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)  

v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}

Here, initial velocity is the final velocity from the first stage. Therefore:  

v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s

3. Use conservation of momentum to find the combined speed of Gayle and her brother.  

Given:

Initial velocity of Gayle and sled is, u_1(i)=10.5 m/s

Initial velocity of her brother is, u_2(i)=0 m/s

Mass of Gayle and sled is, m_1=55.0 kg

Mass of her brother is, m_2=30.0 kg

Final combined velocity is given as:

v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}  

v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79 m/s  

4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.

Using conservation of energy, the final velocity at the bottom of the hill is:  

E(i) = E(f)  

KE(i) + PE(i) = KE(f) + PE(f)  

0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)  

v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s

6 0
3 years ago
A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.460 Hz. The pendulum ha
zlopas [31]

Answer:

The  moment of inertia is  I =1.0697 \ kg m^2

Explanation:

From the question we are told that

    The  frequency is  f  =  0.460 \ Hz

    The  mass of the pendulum is  m  =  2.40  \ kg

    The  location of the pivot from the center is d  =  0.380 \ m

     

Generally the period of the simple harmonic motion is mathematically represented as

        T   = 2 \pi  *  \sqrt{  \frac{I}{ m  *  g *  d  } }

Where I is the moment of inertia about the pivot point , so making I the subject of the formula it

=>    I =  [ \frac{T}{2 \pi } ]^2 *  m*  g * d

But the period of this simple harmonic motion can also be represented mathematically as

        T  =  \frac{1}{f}

substituting values

      T  =  \frac{1}{0.460}

      T  =  2.174 \ s

So

      I =  [ \frac{2.174}{2 * 3.142 } ]^2 *   2.40*  9.8 * 0.380

      I =1.0697 \ kg m^2

4 0
3 years ago
Republicans are called the Liberal Party while the Democrats are called Conservative Party.
Fudgin [204]

Answer:

Explanation:

False

5 0
3 years ago
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When describing the interaction between magnetic poles we could say that
kifflom [539]
The Answer Is A 
Hope This Helps !

8 0
3 years ago
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