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slavikrds [6]
2 years ago
11

What is the gravitational force between two students, John and Mike, if John has a mass of 81.0 kg, Mike has a mass of 93.0 kg,

and their centers are separated by a distance of .620 m?
Physics
1 answer:
Marianna [84]2 years ago
8 0

Answer:

1.31×10¯⁶ N

Explanation:

From the question given above, the following data were obtained:

Mass of John (M₁) = 81 Kg

Mass of Mike (M₂) = 93 Kg

Distance apart (r) = 0.620 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force (F) =?

The gravitational force between the two students, John and Mike, can be obtained as follow:

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 81 × 93 / 0.62²

F = 6.67×10¯¹¹ × 7533 / 0.3844

F = 1.31×10¯⁶ N

Therefore, the gravitational force between the two students, John and Mike, is 1.31×10¯⁶ N

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Hi, thank you for posting your question herein Brainly.

These physical changes could be classified based on their energy requirements: endothermic or exothermic. Endothermic reaction need to absorb energy, while exothermic reaction need to release the energy in order to achieve spontaneous reactions.

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7 0
3 years ago
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A man climbs on to a wall that is 3.6m high and gains 2268J of potential energy. What is the mass of the man? *
Ilia_Sergeevich [38]
The mass of the man is 63 kilograms
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3 0
3 years ago
A 40-cmcm-long tube has a 40-cmcm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. A
Licemer1 [7]

Answer:

1070 Hz

Explanation:

First, I should point out there might be a typo in the question or the question has inconsistent values. If the tube is 40 cm long, standing waves cannot be produced at 42.5 cm and 58.5 cm lengths. I assume the length is more than the value in the question then. Under this assumption, we proceed as below:

The insert in the tube creates a closed pipe with one end open and the other closed. For a closed pipe, the difference between successive resonances is a half wavelength \frac{\lambda}{2}.

Hence, we have

\dfrac{\lambda}{2}=58.5-42.5=16 \text{ cm}

\lambda=32\text{ cm}=0.32 \text{ m}.

The speed of a wave is the product of its wavelength and its frequency.

v=f\lambda

f=\dfrac{v}{\lambda}

f=\dfrac{343}{0.32}=1070 \text{ Hz}

7 0
3 years ago
A tin can collapses if all air inside it is taken out why
Veseljchak [2.6K]

That only happens when the tin can is IN air.

In the familiar, comfy part of Earth's atmosphere where we live, the normal pressure of air is around 14.6 pounds on every square inch of everything. That's a big part of the reason why we're built with bodies that generate that same amount of pressure on the INSIDE pressing OUT. That way, we always have the same pressure pushing in both directions, so we know that we won't get crushed or blow up like balloons.

But we have to be careful with our bodies or other things when they're in places where the atmospheric pressure on the outside is NOT normal.

-- When a deep-sea diver goes hundreds of feet down in the ocean, and the pressure of the water is much GREATER than normal air.

-- When an astronaut has to go outside ... where there's NO air ... and fix something on the International Space Station.

When the pressure on the outside becomes very unusual, we have to wear special suits to protect our bodies from the unusual conditions.

The tin can in the story is a lot like our bodies. As long as it has air inside and air outside, the pressure is the same in both directions, so there's no particular force trying to deform the can. But ...

-- If you seal the can with the air inside it, take the can into a vacuum chamber, and pump the air out of the vacuum chamber, then the can only has pressure inside. It'll expand, and eventually spring a little hole in the metal, and all the air inside will blow out.

-- If you take all the air OUT of the can (so the can is REALLY 'empty'), then the pressure on it is all from the outside. In that situation, the can simply collapses, because there's nothing inside to provide pressure in the outward direction.

One more little thing to think about:

When you want some toothpaste to come drizzling out of the tube onto your brush, what do you do ? Do you perhaps squeeze the tube, and increase the pressure on the outside ?

4 0
3 years ago
A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is pre
Alisiya [41]

Answer:

0.546 \hat k

Explanation:

From the given information:

The force on a given current-carrying conductor is:

F = I ( \L  \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})

where the length usually in negative (x) direction can be computed as

\L ^ {\to }  = -x\hat i \\dL\limits ^ {\to }- dx\hat i

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})

F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)

F = I \int^3_1  - 9.0x^2 \ dx \hat k

F = I  (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k

F = I  (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ]  \hat k

where;

current I = 7.0 A

F = (7.0 \ A)  (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ]  \hat k

F = (7.0 \ A)  (9.0) \bigg [\dfrac{26}{3} \bigg ]  \hat k

F = 546 × 10⁻³ T/mT \hat k

F = 0.546 \hat k

4 0
3 years ago
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