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poizon [28]
3 years ago
11

What is the fundamental source of all energy in the universe

Physics
1 answer:
seropon [69]3 years ago
5 0
The Sun is the world's main source of energy. Without it we would not be able to live.
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The graph shows a ball rolling from A to G. At which point does the ball have the greatest kinetic energy?
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The Earths magnetic field deflects the flow of current from?
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<span>Most of the earth's fresh water is stored as ice in the Arctic and Antarctic regions of the globe.</span>
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3 years ago
Two objects have the same momentum. Do the velocities of these objects necessarily have (a) the same directions and (b) the same
NeX [460]

Answer:

(a) They must have same direction

(b) It is not necessary for them to have same magnitudes

Explanation:

(a)

Momentum is a vector quantity. It is the product of mass (scalar) and velocity (vector). Thus, if the direction of velocity is changed, then as a result the direction of momentum will also change or its magnitude or component in the same direction will change. Hence, for the two objects to have same momentum, the directions of their velocities must also be the same.

(b)

Since, the momentum is product of velocity and mass. It is possible that two bodies of different masses with different velocities might have same momentum, provided the direction of their velocities is same.

For example, take a body of mass 4 kg moving with speed 5 m/s. It will have a momentum of 20 N.s. Now, consider another body of mass 2 kg, moving with speed 10 m/s. It will also have a momentum of 20 N.s.

Thus, it is not necessary for two objects to have same magnitude of velocity to have same momentum.

3 0
3 years ago
A plastic rod has been bent into a circle of radius R = 8.00 cm. It has a charge Q1 = +2.70 pC uniformly distributed along one-q
vovikov84 [41]

Answer:

Part a)

V = -1.52 V

Part b)

V = -1.16 V

Explanation:

Part a)

Electric potential is a scalar quantity

so here we can say that total potential due to a ring on its center is given as

V = \frac{kQ}{R}

here we know that

Q = Q_1 - 6Q_1

Q = - 5 Q_1

Q_1 = 2.70 pC

now we have

V = \frac{(9\times 10^9)(-5\times 2.70 \times 10^{-12})}{0.08}

V = -1.52 V

Part b)

Potential on the axis of the ring is given as

V = \frac{kQ}{\sqrt{r^2 + R^2}}

V = \frac{(9\times 10^9)(-5\times 2.70\times 10^{-12})}{\sqrt{0.08^2 + 0.0671^2}}

V = -1.16 V

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2 years ago
How can Newton's third law describe the forces affecting a rocket as it
rusak2 [61]

Answer:D

Explanation:

Apex

7 0
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