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larisa86 [58]
3 years ago
5

Does the magnetic north pole move every 37 days

Physics
1 answer:
brilliants [131]3 years ago
4 0
No, the magnetic North Pole doesn't move every 37 days. I believe.
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Recently, astronomers have observed stars and other objects that orbit the center of the Milky Way Galaxy farther out than our S
Lesechka [4]

Answer:

That scenario can be explained by the idea of the contribution of dark matter on that point.

Explanation:

It can be explained through the idea of dark matter, this one was born to explain why stars (or any object) that were farther for the supermassive black hole in the center of the Milky Way galaxy didn't decrease it rotational velocity as it was expected according to equation 1.

v = \sqrt{\frac{G M}{r}}  (1)

Where v is the rotational velocity, G is the gravitational constant, M is the mass of the supermassive black hole, and r is the orbital radius.

Notice, that If the distance increases the orbital speed decreases (inversely proportional).

7 0
3 years ago
Meeta used an elastic tape to measure the length of her window to stitch a curtain. Do you think she will be able to stitch a cu
Yuki888 [10]

Answer:

No

Explanation:

She will not be able to measure the length of her window accurately due to instrumental error from her choice of instrument. The elastic nature of her tape would alter the measurement because it will stretch as she is taking her readings, thus reducing the true measurement of the length of her window.

To measure the length of her window, she could use an inelastic tape rule or a metre rule. These instruments would eliminate instrumental error.

3 0
3 years ago
A rocket consists of a shuttle and a fuel tank. The rocket flies through the air with a momentum of 180000kgm/s at a velocity of
babunello [35]

Answer:

Answer is 1000kg

Hope it help

6 0
2 years ago
Use the right-hand rule for magnetic force to determine
Sphinxa [80]
The charge on the moving particle
5 0
3 years ago
A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on the horizontal section of a
seraphim [82]

Answer:

2.38 m/s, 4.31 m/s, lower

Explanation:

a)

Initial energy = final energy

½ m v₀² + ½ I ω₀² = mgh + ½ m v₁² + ½ I ω₁²

Since the ball is rolling without slipping, ω = v / r.

For a hollow sphere, I = ⅔ m r².

½ m v₀² + ½ (⅔ m r²) (v₀ / r)² = mgh + ½ m v₁² + ½ (⅔ m r²) (v₁ / r)²

½ m v₀² + ⅓ m v₀² = mgh + ½ m v₁² + ⅓ m v₁²

⅚ m v₀² = mgh + ⅚ m v₁²

⅚ v₀² = gh + ⅚ v₁²

v₀² = 1.2gh + v₁²

v₁ = √(v₀² − 1.2gh)

Given v₀ = 4.03 m/s, g = 9.80 m/s, h = 0.900 m:

v₁ = √((4.03)² − 1.2 (9.80) (0.900))

v₁ ≈ 2.38 m/s

At the top of the loop, the sum of the forces in the radial direction is:

∑F = ma

W + N = m v² / R

N = m v² / R - mg

N = m (v² / R - g)

Given v = 2.38 m/s, R = 0.450 m, and g = 9.80 m/s²:

N = m ((2.38)² / 0.450 - 9.80)

N = 2.77m

N ≥ 0, so the ball stays on the track.

b)

Initial energy = final energy

Borrowing from part a):

v₂ = √(v₀² − 1.2gh)

This time, h = -0.200 m:

v₂ = √((4.03)² − 1.2 (9.80) (-0.200))

v₂ ≈ 4.31 m/s

c)

Without the rotational energy:

½ m v₀² = mgh + ½ m v₁²

½ v₀² = gh + ½ v₁²

v₀² = 2gh + v₁²

v₁ = √(v₀² - 2gh)

This is less than v₁ we calculated earlier.

6 0
3 years ago
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