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irga5000 [103]
3 years ago
6

An electric wheel chair is designed to run on a single 12-V battery rated to provide 100 ampere-hours. How much energy is stored

in this battery?
Physics
1 answer:
Paha777 [63]3 years ago
5 0

Explanation:

It is given that,

Voltage of the battery, V = 12 V

Current, I = 100 ampere-hours

Energy stored is given by the product of power and time taken. So,

E=P\times t

P is the power, P=V\times I

P=12\times 100

P = 1200 watts

This power can be used for 1 hour or 3600 seconds

Energy, E=1200\times 3600

E = 4320000 J

So, the energy stored in this battery is 4320000 J. Hence, this is the required solution.

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Inez uses hairspray on her hair each morning before going to school. The spray spreads out before reaching her hair partly becau
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The charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Similar charges repel each other, whereas charges that are opposed attract each other.

Given data;

Electric force,F = 9 × 10 ⁻⁹ N

Distance between charges,d = 7 × 10⁻⁴ m

Chrge,q₁ = q₂ =q C

From Columb's law;

\rm F = K \frac{q_1q_2}{d^2} \\\\ 9 \times 10^{-9}  = 9 \times 10^9 \frac{q^2}{(7 \times 10^{-4})^2} \\\\ q^2 = 4.9 \times 10^{-25} \\\\  q = 7 \times 10^{-13} \ C

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2 years ago
A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap
viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

V_r + V_L + V_c = V_{net}

now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

c_2 = -4c_1

c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

5 0
3 years ago
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