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3 years ago

Question 1

Physics

1 answer:

Yuki888 [10]3 years ago

4 0

Answer:

2.Weather patterns are similar along the coast of a continent but different when moving from the coast to the center of a continent.

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An infinitely long wire carrying a 25-A current in a the positive x-direction is places along the xaxis in the vicinity of a 20-

Masteriza [31]

Answer:

0.2 A, clockwise direction

Explanation:

We are given that

Current,I=25 A

Number of turns=n=20

Magnetic field at center of r the loop=B=0

d=2 m

We have to find the direction and magnitude of current flowing in the loop.

Magnetic field due to current I

$B=\frac{\mu_0 I}{2\pi d}\hat{z}$

Magnetic field due to I'

$B'=-\frac{\mu_0 nI''}{2a}\hat{z}$

a=1 m

Net magnetic is zero

Therefore, $B+B'=0$

$\frac{\mu_0 I}{2\pi d}\hat{z}-\frac{\mu_0 nI''}{2a}\hat{z}=0$

$\frac{\mu_0 I}{2\pi d}\hat{z}=\frac{\mu_0 nI''}{2a}\hat{z}$

$\frac{I}{\pi d}=\frac{I'n}{a}$

$I'=\frac{aI}{\pi d n}=\frac{25\times 1}{3.14\times 2\times 20}=0.2 A$

Where $\pi=3.14$

Direction: Clockwise

8 0

3 years ago

How should measurements of car 2 be taken to accurately measure the distance it travels? Be sure to include where the meter stic

faltersainse [42]

Yes it is you are doing such a great job I just needed to tell u you will pass

7 0

3 years ago

If 10. joules of work must be done to move 2.0 coulombs of charge from point A to point B in an electric field, the potential di

masya89 [10]

The equation for work (W) done by an electric field is:

W = qΔV

where q is the magnitude of the charge and ΔV is the potential difference. The question gives you W and q, so plug n' play to find ΔV:

10 = 2ΔV
ΔV = 5

8 0

3 years ago

What is the maximum speed at which a car can safely travel around a circular track of radius 55.0 m If the coefficient of fricti

Stella [2.4K]

Answer:

The maximum speed of the car should be 13.7 m/s

Explanation:

For the car to travel at a maximum safe speed , the frictional force acting should be maximum and at the same time should provide the necessary centripetal force.

Let 'k' (=0.3502) be the coefficient of friction and 'N' be the normal force acting on the surface.

Then ,

N = mg , where 'm' is the mass of the body and 'g'(=9.8) is the acceleration due to gravity.

∴ Maximum frictional force , f = kN = kmg

Centripetal force that should act on the car to move with maximum possible speed is -

$F = \frac{mv^{2} }{r}$ , where 'v' is the velocity of the car and 'r'(=55m) is the radius of circular path.