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Wittaler [7]
3 years ago
8

One wire carries a current of I1= 2 Amps, and the other carries a parallel current (same direction, wires are side by side) of I

2=0.75 Amps. They are 5 meters long in this configuration and separated from each other by 5 cm. What is the magnitude of the force experienced by the wire carried I2?
Physics
1 answer:
Yuki888 [10]3 years ago
6 0

Answer:F=3\times 10^{-5} N

Explanation:

Given

I_1=2 Amps

I_2=0.75 Amps

Length of each wires\left ( L\right )=5 m

Distance between wires \left ( r\right )=5 cm

Force per unit length =\frac{\mu_0I_1I_2}{2\pi r}

F'=\frac{2\times 10^{-7}\times 2\times 0.75}{2\pi \times 0.05}

F'=6\times 10^{-6}

Force for L=5 m

F=3\times 10^{-5} N

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100 g of liquid nitrogen at its boiling point of 77 K is stirred into a beaker containing 500 g of 15°C water.
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Answer:

Option A: none

Explanation:

Specific heat capacity is the amount of heat energy required to raise the temperature of a substance per unit of mass.

From the question, the parameters given are; The heat of vaporization of nitrogen= 48 cal/g and that of water is 80 cal/g.

Using the formulae; Specific heat capacity,c= Q/ m× ∆T,----------–-----------------------------------------------------

STEP ONE: We will have to calculate all the energy numbers; 77k is approximately the boiling point of Nitrogen.

Energy required to decrease water from 15°C to 0°C = E(1).

0.1×10^3 g× 48 cal/gram..

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Energy require to vaporize Nitrogen=E(2).

= 80 cal per gram×0.15(15-0)

= 180 cal

Energy required to decrease water from 15°C to 0°C is higher than that of the energy to vaporize Nitrogen, N2.

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A uniform, solid, 1800.0 kgkg sphere has a radius of 5.00 mm. Find the gravitational force this sphere exerts on a 2.30 kgkg poi
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Answer

given,

Mass of the solid sphere = 1800 Kg

radius of the sphere,R = 5 m

mass of the small sphere, m = 2.30 Kg

when the Point is outside the sphere the Force between them is equal to

      F = \dfrac{GMm}{r^2}   when r>R

When Point is inside the Sphere

      F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}  when r<R

where r is the distance where the point mass is placed form the center

Now Force calculation

a) r = 5.05 m

       F = \dfrac{GMm}{r^2}[/tex]

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b) r = 2.65 m

      F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}

      F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}\ \dfrac{2.65}{5.05}

     F = 5.68\times 10^{-9}\ N

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3 years ago
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