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Natali5045456 [20]
2 years ago
5

In 2014, about how far in meters would you have to travel on the surface of the Earth from the North Magnetic Pole to the Geogra

phic North Pole? Take the radius of the Earth to be 6378.1 km.
Physics
1 answer:
pogonyaev2 years ago
3 0

Answer:

267.07 km

Explanation:

We have given the radius of the earth = 6378.1 km

In 2014 the difference between the magnetic north pole and geographical north pole is 2.40°

2.40°=2.40\times \frac{\pi }{180}=0.041866 radian

We know that linear distance is given by S=R\Theta =6378.1\times 0.041866=267.07km

So we have to travel 267.07 km in going from magnetic north pole to geographic north pole

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Answer:

THE FIRST ONE YOU SHOULD TELL HIM AND THE LAST ONE YOU SHOUDENT DO BECAUSE HE WILL DO IT AGAIN AND EXPECT OTHERS TO CLEAN UP AFTER HIM

Explanation:

5 0
1 year ago
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Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p
worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

  • U1/U2=6
  • U1/U2=6

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy  is mathematically given as

U=\frac{-k*qQ}{r}

Therefore

The Equations of U1 and U2 is

For U1

U1=\frac{-k*q_1Q}{r}

For U2

U2=\frac{-k*q_1Q}{3*2r}

Since

U is a function of q and  q2=q1/3

Therefore

U1/U2=6

For Question 2

For U1

U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}

Therefore

U1/U2=6

For more information on this visit

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7 0
1 year ago
g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobb
Law Incorporation [45]

Answer:

Explanation:

total weight acting downwards

= 3g + 10g

13 g

volume of lead = 10 / 11.3 = .885 cm³

Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber  = v x 1 x g

Buoyant force on lead =  .885 x 1 x g

total buoyant force = vg + .885 g

For floating

vg + .885 g  = 13 g

v = 12.115 cm³

total volume of bobber

= 4/3 x 3.14 x 2³

= 33.5 cm³

fraction of volume submerged

= 12.115  / 33.5

= .36  

= 36 %

4 0
1 year ago
Light from a monochromatic source shines through a double slit onto a screen 5.00 m away. The slits are 0.180 mm apart. The dark
Nitella [24]

Answer:

Wavelength of incident light, \lambda = 612 nm

Given:

Distance between slit and screen, x = 5.00 m

slit width, d = 0.180 mm

width of the fringe, \beta = 1.70 cm = 0.017 m

Solution:

To calculate the wavelength of the incident light, \lambda:

\beta = \frac{x\lambda }{d}

\lambda = \frac{\beta d}{x}

\lambda = \frac{0.017\times 0.180\times 10^{- 3}}{5} = 6.12\times 10^{- 7}m = 612 nm

\lambda = 612 nm

4 0
1 year ago
Select the correct answer. Which magnet will produce the strongest force of attraction when placed beneath this magnet? A. B. C.
Brut [27]

The answer to this question is B

4 0
1 year ago
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