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2 years ago

A closed curve encircles several conductors. The line integral around this curve is (image attached below)

Physics

1 answer:

Masteriza [31]2 years ago

5 0

The net current in the conductors and the value of the line integral

$I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A$
The resultant remains same 3.2 *10^4 Tm

This is further explained below.

<h3>What is the net current in the conductors?</h3>

Generally,

To put it another way, the total current In flowing across a surface S (contained by C) is proportional to the line integral of the magnetic B-field (in tesla, T).

$\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \iint_S \mathbf{J} \cdot \mathrm{d}\mathbf{S} = \mu_0I_\mathrm{enc}$

$I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A$

In conclusion, It is possible for the line integral to go around the loop in either direction (clockwise or counterclockwise), the vector area dS to point in either of the two normal directions and Ienc, which is the net current passing through the surface S, to be positive in either direction—but both directions can be chosen as positive in this example. The right-hand rule solves these ambiguities.

The resultant remains the same at 3.2 *10^4 Tm

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In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

VLD [36.1K]

Answer:

(a) 8.2 x 10^-8 N

(b) 3.6 x 10^-47 N , 2.27 x 10^39

Explanation:

charge of proton, q1 = 1.6 x 10^-19 C

charge of electron, q2 = - 1.6 x 10^-19 C

radius of orbit, r = 0.053 nm = 0.053 x 10^-9 m

mass of electron, me = 9.1 x 10^-31 kg

mass of proton, mp = 1.67 x 10^-27 kg

Gravitational constant, G = 6.67 x 10^-11 Nm^2/kg^2

(a) The electrostatic force between two charges is given by

$F_{e}=\frac{Kq_{1}q_{2}}{r^2}$

Where, K is the coulombic constant = 9 x 10^9 Nm^2/C^2

By substituting the values

$F_{e}=\frac{9\times 10^{9}\times 1.6\times 10^{-19}\1.6\times 10^{-19}}{\left ( 0.053 \times 10^{-9} \right )^2}$

Fe = 8.2 x 10^-8 N

(b) The gravitational force between the electron and proton is given by

$F_{g}=\frac{Gm_{e}m_{p}}{r^{2}}$

$F_{g}=\frac{6.67\times 10^{-11}\times 9.1\times 10^{-31}\1.67\times 10^{-27}}{\left ( 0.053 \times 10^{-9} \right )^2}$

Fg = 3.6 x 10^-47 N

$\frac{F_{e}}{F_{g}}=\frac{8.2\times10^{-8}}{3.6\times 10^{-47}}$

$\frac{F_{e}}{F_{g}}=2.27\times 10^{39}$

5 0

3 years ago

Approximately 80% of the energy used by the body must be dissipated thermally. The mechanisms available to eliminate this energy

Semenov [28]

Answer:

the correct answer is c) 23 g

Explanation:

The heat lost by the runner has two parts: the heat absorbed by sweat in evaporation and the heat given off by the body

Q_lost = - Q_absorbed

The latent heat is

Q_absorbed = m L

The heat given by the body

Q_lost = M $c_{e}$ ΔT

where m is the mass of sweat and M is the mass of the body

m L = M c_{e} ΔT

m = M c_{e} ΔT / L

let's replace

m = 90 3.500 1.8 / 2.42 10⁶

m = 0.2343 kg

reduced to grams

m = 0.2342 kg (1000g / 1kg)

m = 23.42 g

the correct answer is c) 23 g

8 0

3 years ago

What is the largest possible displacement resulting from two displacements with magnitudes 3m and 4m? what is the smallest possi

Julli [10]

The largest possible displacement you can get by combining the two of them is if they're both in the SAME direction. You walk 3 meters, stop for a breath, then walk another 4 meters in the same direction as the first one. The resultant is 7 meters.

The smallest possible resultant is 1 meter, if the two displacements are in OPPOSITE directions. This time, you walk 4 meters, stop for a breath, then you TURN AROUND and walk 3 meters back in the direction you came from. You end up 1 meter from where you started.

4 0

3 years ago

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Why is the reactor coolant water kept contained within the primary loop instead of allowing it to mix with the feed water and le

uysha [10]

For the answer to the question above, I the answer is yes, It is both contaminated and radioactive at the same time. That's why they keep the water spinning.I hope my answer helped you. Have a nice day!

4 0

3 years ago

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A freight car of mass M contains a mass of sand m. At t = 0 a constant horizontal force F is applied in the direction of rolling

Veronika [31]

Answer:

Amount of linear movement

Explanation:

Our system is defined by the rate of change in mass that

leaves the car $\Delta m_ {s}$ , this happens during a time interval