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3 years ago

Why does a compass point towards earths geographic North Pole

Physics

2 answers:

kakasveta [241]3 years ago

6 0

Answer:

2) Earth’s geographic North Pole is close to the magnetic south pole.

3) Magnets allowed to move freely align themselves to point north.

5) The north pole on a compass is attracted to Earth’s magnetic south pole.

Hope this helped:)

Travka [436]3 years ago

3 0

It doesn't. It points to the Earth's MAGNETIC pole. There's only a very small part of the Earth's surface where the geographic pole is in exactly the same direction, and there are places where there's a huge difference. For example, I've worked in parts of Alaska where compasses point WEST !

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The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

miss Akunina [59]

Formula of the gravitational force between two particles:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67 \cdot 10^{-11} Nm^2 kg^{-2}$ is the gravitational constant

m1 and m2 are the masses of the two particles

r is their distance

(a) particle A

The gravitational force exerted by particle B on particle A is

$F_B=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=5.14 \cdot 10^{-5} N$ to the right

The gravitational force exerted by particle C on particle A is

$F_C=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=5.6 \cdot 10^{-6} N$ to the right

So the net gravitational force on particle A is

$F_A = F_B + F_C =5.14 \cdot 10^{-5} N+5.6 \cdot 10^{-6} N=5.7 \cdot 10^{-5} N$ to the right

(b) Particle B

The gravitational force exerted by particle A on particle B is

$F_A=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=-5.14 \cdot 10^{-5} N$ to the left

The gravitational force exerted by particle C on particle B is

$F_C=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=8.41 \cdot 10^{-5} N$ to the right

So the net gravitational force on particle B is

$F_B = F_A + F_C =-5.14 \cdot 10^{-5} N+8.41 \cdot 10^{-5} N=3.27 \cdot 10^{-5} N$ to the right

The gravitational force exerted by particle A on particle C is

$F_A=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=-5.6 \cdot 10^{-6} N$ to the left

The gravitational force exerted by particle B on particle C is

$F_B=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=-8.41 \cdot 10^{-5} N$ to the left

So the net gravitational force on particle C is

$F_C = F_B + F_A =-8.41 \cdot 10^{-5} N-5.6 \cdot 10^{-6} N=-8.97 \cdot 10^{-5} N$ to the left

3 0

3 years ago

A measurement standard is defined as ____.

Serggg [28]

Answer:

A measurement standard is defined as 'the exact quantity people agree to use for comparison'.

Explanation:

For the same type of quantity, there is a unit of measurement which is defined and adopted either conventionally or by law which is used as standard for measurement of that quantity.

This measurement standard is defined as 'the exact quantity people agree to use for comparison'.

For example, for length the standard unit of measurement is 'meter'.

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lina2011 [118]

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What happens when heat is removed from water?

IgorLugansk [536]

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Tom [10]

External force.

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If there is no external force on a moving object, it keeps going.

Forever !

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