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3 years ago

9

what is the net force on an object that is experiencing a force of 25 N north, a force of 25 N south, a force of 50 N to the eas

t and a force of 45 N to th west?

1 answer:

REY [17]3 years ago

4 0

Answer:

5 n

Explanation:

25 and 25 cancel each other out and 50-45 is 5

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in bohr's atomic theory when electron moves from one energy level to another energy level more distant from the nucleus energy i

Katyanochek1 [597]

No, it is the other way around. When an electron moves from one energy level to another energy level more distant from the nucleus, it gains, not emits energy. The closer it gets to the nucleus, the more energy it emits. If it is far from the nucleus, it gains more energy.

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3 years ago

If a car has a velocity of 40 km/h west and then comes to a stop in 5 s, what is its acceleration in m/s2?

Leno4ka [110]

acceleration = change in velocity /change in time

convert 40km to meter then divide it with 5

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3 years ago

If a point has 40 J of energy and the electric potential is 8 V, what must be the charge?

Alekssandra [29.7K]

If a point has 40 J of energy and the electric potential is 8 V, the charge must be: A. 5 C

<u>Given the following the details;</u>

Energy = 40 Joules

Electric potential = 8 Volts

To find the quantity of charge;

Mathematically, the quantity of charge with respect to electric potential is given by the formula;

$Quantity \; of \; charge = \frac{Energy}{Electric \; potential}$

Substituting the values into the formula, we have;

$Quantity \; of \; charge = \frac{40}{8}$

<em>Quantity of charge = 5 Coulombs</em>

Therefore, the quantity of charge must be <em>5 Coulombs.</em>

Find more information: brainly.com/question/21808222

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3 years ago

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The Sun has a mass of 1.99x10^30 kg and a radius of 6.96x10^8 m. Calculate the acceleration due to gravity, in meters per second

just olya [345]

Answer:

$g=274\ m/s^2$

Explanation:

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

The radius of the Sun, $r=6.96\times 10^8\ m$

We need to find the acceleration due to gravity on the surface of the Sun. It is given by the formula as follows :

$g=\dfrac{GM}{r^2}\\\\g=\dfrac{6.67\times 10^{-11}\times 1.99\times 10^{30}}{(6.96\times 10^8)^2}\\\\g=274\ m/s^2$

So, the value of acceleration due to gravity on the Sun is $274\ m/s^2$ .

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3 years ago

What are the opposite ends of a magnet called? A. north and south terminals B. north and south poles C. magnetic fields D. magne

Bezzdna [24]

I think the answer is B.

Hope this helps.

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3 years ago

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