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IrinaVladis [17]
3 years ago
13

If the strength of the magnetic field at B is 3 units, the strength of the magnetic field at A is ____.

Physics
2 answers:
anyanavicka [17]3 years ago
3 0

I think it is C. 12 units please tell me if im wrong

AfilCa [17]3 years ago
3 0

Answer:

Correct Answer choice is C) 12! Hope this helped!

Explanation:

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7 0
2 years ago
Question 1 of 25
finlep [7]

Answer:

<em>2.753*10^-11N</em>

Explanation:

According to Newton's law of gravitation, the force between the masses is expressed as;

F = GMm/d²

M and m are the distances

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Given

M = 3.71 x 10 kg

m = 1.88 x 10^4 kg

d = 1300m

G = 6.67 x 10-11 Nm²/kg

Substitute into the formula

F = 6.67 x 10-11* (3.71 x 10)*(1.88 x 10^4)/1300²

F = 46.52*10^(-6)/1.69 * 10^6

F = 27.53 * 10^{-6-6}

F = 27.53*10^{-12}

F = 2.753*10^-11

<em>Hence the gravitational force between the asteroid is 2.753*10^-11N</em>

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6 0
2 years ago
Anyone please help me with this question ASAP
Katyanochek1 [597]

Answer:

increases

Explanation:

it would have to work harder to get to two points together

7 0
3 years ago
A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between
SashulF [63]

To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:

\Delta W = \Delta KE

\Delta W = \frac{1}{2} mv^2

Here,

m = mass

v = Velocity

Our values are given as,

m = 79.7kg

v = 4.77m/s

Replacing,

\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2

\Delta W = 907J

Therefore the mechanical energy lost due to friction acting on the runner is 907J

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3 years ago
In 1998, scientists discovered that the expansion of the universe has been accelerating.
Gnom [1K]

Answer:the answer should be dark energy

Explanation:

3 0
3 years ago
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