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3 years ago

If 3.2*10^20 electrons pass through a wire in 4s, what would be the electrical current in the wire?

Physics

1 answer:

rewona [7]3 years ago

4 0

Answer: 12.8 A

Explanation:

Current is defined as the rate of flow of electric charge based on the formula:

I(current) = deltaQ(change in charge)/deltat(change in time).

First, however, we must convert the number of electrons into the number of coulombs. Based on the fact that the charge of 1 electron or 1 elementary charge is equal to 1.6*10^-19 C, we can calculate:

3.2*10^20 e = 1.6*10^-19*3*10^20 C = 51.2 C.

Now we use: I = Q/t = 51.2/4 = 12.8 A.

Hope this helped.

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Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 18.0 cm and carries a

Helen [10]

Answer:

Explanation:

The wires are in circular shape . They have common center .

magnetic field due to circular wire is given by the formula

B = $\frac{\mu_0\times I }{2r}$

where I is current , r is radius of the coil .

magnetic field due to inner wire

= $\frac{\mu_0\times 10 }{2\times.09}$

magnetic field due to outer wire

= $\frac{\mu_0\times I }{2\times.15}$

These should be equal and opposite so that by cancelling each other , they create zero field.

$\frac{\mu_0\times 10 }{2\times.09}$ = $\frac{\mu_0\times I }{2\times.15}$

I = 16.66 A

Direction of current should be in opposite direction ie anticlockwise when looking from above.

6 0

3 years ago

In this unit, the amount of solute is measured in?

Hoochie [10]

Try liters if you haven’t done it yet. I’m so sorry if i’m incorrect.

5 0

3 years ago

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Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the

Masja [62]

Answer:

distance of 2nd team from 1st team will be: 58.2

Direction of 2nd team from 1st team will be: 14.90 deg North of east

Explanation:

ASSUME Vector is R and makes angle A with +x-axis,

therefore component of vector R is

$R_x = Rcos A$

$R_y = Rsin A$

From above relation

Assuming base camp as the origin, location of 1st team is

$R_1 = 37 km$ away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

$R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km$

$R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km$

location of 2nd team is at

$R_2 = 32 km$ , at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

$R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km$

$R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km$

Now position of 2nd team with respect to 1st team will be given by:

$R_3 = R_2 - R_1$

$R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j$

Using above values:

$R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j$

$R_3 = 51.49 i + 13.71 j$

distance of 2nd team from 1st team will be:

$\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)$

$\left | R_3 \right | = 53.28 km = 58.2 km$

Direction of 2nd team from 1st team will be:

$Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]$

Direction = 14.90 deg North of east

6 0

3 years ago

A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls

nekit [7.7K]

Given:

density of air at inlet, $\rho_{a} = 1.20 kg/m_{3}$

density of air at inlet, $\rho_{b} = 1.05 kg/m_{3}$

Solution:

Now,

$\dot{m} = \dot{m_{a}} = \dot{m_{b}}$

$\rho_{a} A v_{a} = \rho _{b} Av_{b}$ (1)

where

A = Area of cross section

$v_{a}$ = velocity of air at inlet

$v_{b}$ = velocity of air at outlet

Now, using eqn (1), we get:

$\frac{v_{b}}{v_{a}} = \frac{\rho_{a}}{\rho_{b}}$

$\frac{v_{b}}{v_{a}} = \frac{1.20}{1.05}$ = 1.14

% increase in velocity = $1.14\times 100$ =114%

which is 14% more

Therefore % increase in velocity is 14%

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3 years ago

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.. As you increase the

Llana [10]

Answer:

air pressure increases and temperature decreases

Explanation:

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3 years ago