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Alex_Xolod [135]
2 years ago
10

Alfredo leaves camp and, using a compass, walks 4 km E, then 6 km S, 3 km E, 5 km N, 10 km W, 8 km N, and, finally, 3 km S. At t

he end of two days, he is planning his trip back. To get back to camp, Alfredo should travel a distance of ----------?
km in a direction
---------------? ° south of east.
Physics
1 answer:
kari74 [83]2 years ago
8 0

Answer: 3km E, 1km N

Explanation:

4E + 6S + 3E + 5N + 10W + 8N + 3S= 3W, 1S

3W, 1S= 3E, 1N

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Based on illustrations of magnetic field lines, where could an object be placed so it would not experience a magnetic force
tiny-mole [99]

<u>Halfway</u><u> between the like poles of two magnets, because the field lines bend away and do not enter this area.</u>

How does a magnetic field diagram show where the field is strongest?

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  • The lines include arrowheads to indicate the direction of the force exerted by a magnetic north pole.
  • The closer the lines are to the poles, the stronger the magnetic field (thus the magnetic field from a bar magnet is highest closest to the poles).

Where is magnetic field the strongest and weakest on a magnet?

  • The bar magnet's magnetic field is strongest at its core and weakest between its two poles.
  • The magnetic field lines are densest immediately outside the bar magnet and least dense in the core.

Which two locations on the magnet would have the greatest attractive forces?

  • Inside the magnet itself, the field lines run from the south pole to the north pole.
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Learn more about magnetic field

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4 0
2 years ago
A large asteroid of mass 98700 kg is at rest far away from any planets or stars. A much smaller asteroid, of mass 780 kg, is in
Citrus2011 [14]

Answer:

1.81 x 10^-4 m/s

Explanation:

M = 98700 kg

m = 780 kg

d = 201 m

Let the speed of second asteroid is v.

The gravitational force between the two asteroids is balanced by the centripetal force on the second asteroid.

\frac{GMm}{d^{2}}=\frac{mv^2}{d}

v=\sqrt{\frac{GM}{d}}

Where, G be the universal gravitational constant.

G = 6.67 x 10^-11 Nm^2/kg^2

v=\sqrt{\frac{6.67 \times 10^{-11}\times 98700}{201}}

v = 1.81 x 10^-4 m/s

7 0
4 years ago
What is the direction of the centripetal force when applied to an object?
mihalych1998 [28]
For an object moving in a path that's a circle or a part of one,
the centripetal force acts in the direction toward the center of
the circle.  That direction is perpendicular to the way the object
is moving.
3 0
3 years ago
Read 2 more answers
We would be more likely to get an accurate measurement of the distance from Earth to a nearby star if we took the angular measur
NeTakaya

Answer:

6 month interval

Explanation:

The distance to a nearby star in theory is more simple than

one might think! First we must learn about the parallax effect. This is the mechanism our eyes use to perceive things at a distance! When we look at the star from the earth we see it at different angles throughout the earth's movement around the sun similar to how we see when we cover on eye at a time. Modern telescopes and technology can help calculate the angle of the star to the earth with just two measurements (attached photo!) Since we know the distance of the earth from the sun we can use a simple trigonometric function to calculate the distance to the star. The two measurements needed to calculate the angle of the star to the earth caused by parallax (in short angle θ) are shown in the second attached photo.

So using a simple trigonometric function Sin\theta=\frac{r}{d} we can solve for d which is the distance of the earth to the star:

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In the first attached photo a picture where r is the distance to the star and the base of the triangle is the diameter of the earth.

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What is potential
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Its b, energy waiting to be used
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