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Damm [24]
3 years ago
9

Suppose a straight 1.00-mm-diameter copper wire could just "float" horizontally in air because of the force due to the Earth’s m

agnetic field which is horizontal, perpendicular to the wire, and of magnitude 5.0x10-5 T. What current would the wire carry? Does the answer seem feasible? Explain briefly.
Physics
1 answer:
insens350 [35]3 years ago
8 0

To solve this problem it is necessary to apply the concepts related to the Force since Newton's second law, as well as the concept of Electromagnetic Force. The relationship of the two equations will allow us to find the magnetic field through the geometric relations of density and volume.

F_{mag}= BIL

Where,

B = Magnetic Field

I = Current

L = Length

<em>Note: F_{mag}  is a direct adaptation of the vector relation F=q \times V \times B</em>

From Newton's second law we know that the relation of Strength and weight is determined as

F_g = mg

Where,

m = Mass

g = Gravitational Acceleration

For there to be balance the two forces must be equal therefore

F_{mag} = F_g

BIL = mg

Our values are given as,

Diameter (d) = 1.0mm = 1*10^{-3}m

Radius (r) = \frac{d}{2} = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m

Magnetic Field (B) = 5.0*10^{-5} T

From the relationship of density another way of expressing mass would be

\rho = \frac{m}{V} \rightarrow m = \rho V

At the same time the volume ratio for a cylinder (the shape of the wire) would be

V = \pi r^2 L \rightarrow L =Length, r= Radius

Replacing this two expression at our first equation we have that:

BIL = mg

BIL = ( \rho V)g

BIL = ( \rho \pi r^2 L)g

Re-arrange to find I

I = \frac{( \rho \pi r^2 L)g}{BL}

I = \frac{( \rho \pi r^2 )g}{B}

We have for definition that the Density of copper is 8.9*10^3 Kg/m^3, gravity acceleration is 9.8m/s^2 and the values of magnetic field (B) and the radius were previously given, then:

I = \frac{( (8.9*10^3 ) \pi (0.5*10^{-3})^2 )(9.8)}{5.0*10^{-5}}

I = 1370.05A

The current is too high to be transported which would make the case not feasible.

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3 years ago
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Juan and Kuri are on a carousel. Juan is closer to the center of the carousel than Kuri. Which statement describes their tangent
Licemer1 [7]

Answer:

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

Explanation:

Since Juan is closer to the center and Kuri is away from the center so we can say that Juan will move smaller distance in one complete revolution

As we know that the distance moved in one revolution is given as

d = 2\pi r

also the time period of revolution for both will remain same as they move with the time period of carousel

Now we can say that the speed is given as

v = \frac{2\pi r}{T}

so Juan will have less tangential speed. so correct answer will be

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

6 0
3 years ago
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It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an
Maru [420]

Answer:

q=7.965*10^-^6C

Explanation:

From the question we are told that

Altitude of  d_1 390m

MagnitudeM_1=60.0 N/C

Altitude of d_2=240 m

Magnitude is M_2= 100 N/C

Distance of cube d_c=150 m

Generally the flux \phi is mathematical given as

 \phi=60(150)^2cos180+100(150)^2*cos0

 \phi=-9*10^5

Generally Quantity of charge  q is mathematically given as

 q=\varepsilon _0 *\phi

 q=8.85*10^-^1^2 *9*10^5

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5 0
3 years ago
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te
schepotkina [342]

Answer:

6.77 minutes

Explanation:

172 degree - 78 degree = (185 degree - 78 degree)e−2 k

=> 94 = 107

e−2 k => 94 ÷ 107

k => ln (94÷107) / 2

147 - 78 = (185 - 78)e ^[ln (94÷107) / 2]

=> 69 = 107 e^ [ln (94÷107) / 2]

e^[ln (94÷107) / 2] =69 ÷ 107

=> t = [ln (69 ÷ 107)] ÷ [ln (94÷107) / 2]

t=> -0.4387 ÷ -0.0648

t => 6.77 minutes.

Therefore, the final answer to the question is 6.77 minutes.

4 0
3 years ago
A person is hauling their taco stand and it takes 3,500 Joules of work to stop the taco stand. What force was exerted on the tac
Lapatulllka [165]

Given the work done, the force exerted on the taco stand through the given distance is 2.3 × 10³ Newtons.

<h3>What is Work done?</h3>

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

W = f × d

Where f is force applied and d is distance travelled.

Given that;

  • Work done W = 3500J = 3500kgm²/s²
  • Distance covered d = 1.5m
  • Force applied F = ?

W = f × d

3500kgm²/s² = f × 1.5m

f = 3500kgm²/s² ÷ 1.5m

f = 2.3 × 10³ kgm/s²

f = 2.3 × 10³ N

Given the work done, the force exerted on the taco stand through the given distance is 2.3 × 10³ Newtons.

Learn more about work done here: brainly.com/question/26115962

#SPJ1

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2 years ago
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