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MArishka [77]
3 years ago
9

Earth's magnetic field is approximately 1/2 gauss, that is 50 micro-tesla because the SI field unit of a tesla is 10,000 gauss.

Earth's north geographic pole is close to its south magnetic pole, and magnetic field is directed from the north to the south poles of a magnetic dipole so it goes from Earth's south geographic pole towards its north. Suppose you have wire carrying a large DC current from the south wall of a building to its north wall and that it is horizontal, on the floor. If Earth's field is parallel to the ground and does not dip, what force if any would the wire experience
Physics
1 answer:
jek_recluse [69]3 years ago
5 0

Answer:

F = 0

Explanation:

The magnetic force is described by two expressions

for a moving charge

          F = q v x B

for a wire with a current

         F = I L xB

bold indicates vectors

let's write this equation in module form

         F = I L B sin θ

where the angle is between the direction of the current and the direction of the magnetic field

In this case they indicate that the cable goes from the South wall to the North wall, so this is the direction of the current

The magnetic field of the Earth goes from the south to the north and in this part it is horizontal

Therefore the current and the magnetic field are parallel, the angle between them is zero

           sin 0 = 0

consequently the magnetic force is zero

            F = 0

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Harman [31]

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Explanation:

The described situation is as follows:

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This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

V_{f}=V_{o}+at (1)  

{V_{f}}^{2}={V_{o}}^{2}+2ad (2)  

Where:  

V_{f} Is the final velocity of the object

V_{o}=0 Is the initial velocity of the object (it was dropped)

a=9.8 m/s^{2} is the acceleration due gravity

d is the height of the tower

t=0.02min=1.2 s is the time it takes to the object to reach the ground

b) Begining with (1):

V_{f}=0+at (3)  

V_{f}=at=(9.8 m/s^{2})(1.2 s) (4)  

V_{f}=11.76 m/s (5)  This is the final velocity of the object

a) Substituting (5) in (2):

(11.76 m/s)^{2}=0+2(9.8 m/s^{2})d (6)  

Clearing d:

d=\frac{(11.76 m/s)^{2}}{2(9.8 m/s^{2})} (7)  

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3 years ago
A distant galaxy is determined to be 150 million light years distant and moving away from us; using the Hubble law determine its
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The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
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We'll also need to know that 1 parsec = about 3.262 light years.

So the speed of your receding galaxy is

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4 0
3 years ago
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