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kvv77 [185]
3 years ago
11

Use the drop-down menus to complete the statements magnetic field lines exit out of the north pole south pole magnetic field lin

es into into the north pole south pole magnetic field lines travel around a bar magnet in a closed loop an open
Physics
2 answers:
MAXImum [283]3 years ago
4 0

Answer:

1. North Pole

2. South Pole

3. A closed loop

Explanation:

V125BC [204]3 years ago
3 0

Answer:

<h2>north pole south pole and a closed loop </h2>

Explanation:

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Which of the following is not a characteristic of a electrical potential energy?
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It's a form of mechanical energy
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Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
MArishka [77]

Complete Question:

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer:

m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

6 0
3 years ago
tyhe mass of an object is 117 g adding 1200j of heat will raise the temperture of the object by 12 celsius what is the specifc h
Sati [7]

Answer:

C 0.85 j/g*k

Explanation:

The specific heat capacity of a material is given by:

C_s = \frac{Q}{m \Delta T}

where

Q is the amount of heat supplied to the object

m is the mass of the object

\Delta T is the increase in temperature of the object

For the object in this problem, we have

m = 117 g is the mass

Q = 1200 J is the heat supplied

\Delta T=12^{\circ} is the increase in temperature

Substituting into the formula, we find the specific heat:

C_s = \frac{1200 J}{(117 g)(12^{\circ})}=0.85 J/gC

8 0
3 years ago
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