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3 years ago

Which of the following statements are true?

Physics

2 answers:

nalin [4]3 years ago

5 0

Answer:

Positively charged objects attract negatively charged objects.

Explanation:

This is due to a law that states 'like forces attract while unlike forces repel. This same concept applies to magnetism. If you put two similar poles together, for example; if you place two south poles together. You feel a separating force between the two poles. But if you place two opposite poles together they attract each other. Hope i helped.

Dmitrij [34]3 years ago

3 0

Answer:

Positively charged objects attract negatively charged objects.

Explanation:

hope this helps

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17. Which of the following is NOT a testable

Artemon [7]

Answer:

c. testing student opinions

Explanation:

opinions aren't factual and they would not aide an experiment if it wasn't for a social experiment.

7 0

2 years ago

Find the magnitude of the free-fall acceleration at the orbit of the Moon (a distance of 60RE from the center of the Earth with

Ede4ka [16]

Answer:

The magnitude of the free-fall acceleration at the orbit of the Moon is $2.728\times 10^{-3}\,\frac{m}{s^{2}}$ ( $\frac{2.784}{10000}\cdot g$ , where $g = 9.8\,\frac{m}{s^{2}}$ ).

Explanation:

According to the Newton's Law of Gravitation, free fall acceleration ( $g$ ), in meters per square second, is directly proportional to the mass of the Earth ( $M$ ), in kilograms, and inversely proportional to the distance from the center of the Earth ( $r$ ), in meters:

$g = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$G$ - Gravitational constant, in cubic meters per kilogram-square second.

$M$ - Mass of the Earth, in kilograms.

$r$ - Distance from the center of the Earth, in meters.

If we know that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972\times 10^{24}\,kg$ and $r = 382.26\times 10^{6}\,m$ , then the free-fall acceleration at the orbit of the Moon is:

$g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(382.26\times 10^{6}\,m)^{2}}$

$g = 2.728\times 10^{-3}\,\frac{m}{s^{2}}$

6 0

3 years ago

Emily holds a banana of mass m over the edge of a bridge of height h. She drops the banana and it falls to the river below. Use

bearhunter [10]

Answer:

The mass of the banana is m and it is at height h.

Applying the Law of Conservation of Energy

Total Energy before fall = Total Energy after fall

$E_{i}$ = $E_{f}$

Here, total energy is the sum of kinetic energy and potential energy

$K.E_{i}$ + $P.E_{i}$ = $K.E_{f}$ + $P.E_{f}$ (a)

When banana is at height h, it has

$K.E_{i}$ = 0 and $P.E_{i}$ = mgh

and when it reaches the river, it has

$K.E_{f}$ = 1/2m $v^{2}$ and $P.E_{f}$ = 0

Putting the values in equation (a)

0 + mgh = 1/2m $v^{2}$ + 0

mgh = 1/2m $v^{2}$

cutting 'm' from both sides

gh = 1/2 $v^{2}$

v = $\sqrt{2gh}$

Hence, the velocity of banana before hitting the water is

v = $\sqrt{2gh}$

5 0

3 years ago

Calculate the force generated by a car that hits the wall at an

Makovka662 [10]

This is a defective question. It was WRITTEN by someone who is unclear on the concepts. DON'T try and answer it.

It's trying to get us to use Newton's second law ... F = m • a.

But that only tells us how much force must act ON THE CAR in order to accelerate it. (45 kg) • (4 m/s^2) = 180 newtons.

This is NOT the force exerted BY the car when it hits something. THAT force depends on its speed WHEN it hits, AND how long it takes for the wreckage to actually come to rest, AND how hard or soft the wall is.

DON'T try to answer this question. Your answer will be wrong, you won't understand why, and the teacher you try to argue with probably won't either.

============================================

More explanation:

Think about jumping off of a ladder in your back yard. Several times.

Your mass is the same every time. Your acceleration is the same every time . . . 9.8 m/s² down, the acceleration of Earth gravity, every time.

BUT ...

-- I'll bet you would rather land on wood than on concrete. The force of landing would be less.

-- I'll bet you would rather land on dirt than on wood. The force of landing would be less.

-- I'll bet you would rather land on grass than on dirt. The force of landing would be less.

-- I'll bet you would rather land on a pile of blankets than on dirt. The force of landing would be less.

-- I'll bet you would rather land on a trampoline than on a pile of blankets. The force of landing would be less.

-- I'll bet you would rather jump from a short ladder than from a tall one. Your speed would be less when you landed, and the force of landing would be less.

==> Your mass is the SAME every time, and your acceleration is the SAME every time. But the force when you hit is DIFFERENT every time.

The mass and acceleration of the car DON'T tell us the force of the hit when the car hits a wall.

6 0

3 years ago

Why do noble gasses rarely react with other elements?

ollegr [7]

Noble gasses have an outer shell full of electrons. A full outer energy level is the most stable arrangement of electrons. As a result, noble gases cannot become more stable by reacting with other elements and gaining or losing valence electrons. Therefore, noble gases are rarely involved in chemical reactions and almost never form compounds with other elements.

3 0

2 years ago