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Marat540 [252]
3 years ago
10

The force per meter between the two wires of a jumper cable being used to start a stalled car is 0.244 N/m. (a) What is the curr

ent (in A) in the wires, given they are separated by 1.70 cm
Physics
1 answer:
MakcuM [25]3 years ago
8 0

Answer: current I is 144amp(A)

Explanation: The force per lenght F/L, current l, and separation distance r are related by the formula

F/L = permeability of space* I²/2*pi*r

Permeability of free space is 1.257*EXP {-6} Henry/meter

F/L is 0.244 N/m, pi is 3.142,

r is 1.7cm which is equal to 0.017meter.

I is what we are looking for.

Now we have after substitution,

0.244

= 1.257*EXP {-6} * I²/(2*3.142*0.017)

After cross multiplying and making I² subject of formula we have,

0.0261/(1.257*EXP {-6}) = I²

I = √{20736.7} =144ampere(A)

That means in each of the jumper wire there is 144amp(A)

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3 years ago
Which season is signaled by average lower temperature and indirect, angled sunlight?
anygoal [31]

When the earth in its translation movement has the earth's axis is further from the sun and the energy per unit area is minimal, we have the winter season

 

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When the rays arrive with the greater inclination, the energy deposited per unit area is minimal, which is why the temperature decreases, this period is called Winter.

In the periods when the axis is almost vertical we have an average absorption of energy, these two periods are called Spring and Autumn.

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4 0
3 years ago
A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is t
sertanlavr [38]

Explanation:

The given data is as follows.

       C = 20 \times 10^{-6} F

        R = 100 \times 10^{3} ohm

        Q_{o} = 100 \times 10^{-6} C

          Q = 13.5 \times 10^{-6} C

Formula to calculate the time is as follows.

          Q_{t}  = Q_{o} [e^{\frac{-t}{\tau}]

       13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]

               0.135 = e^{\frac{-t}{2}}

         e^{\frac{t}{2}} = \frac{1}{0.135}

                         = 7.407

           \frac{t}{2} = ln (7.407)

                      t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

4 0
3 years ago
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