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egoroff_w [7]
2 years ago
10

Might it be possible to explain the interaction of the rod and pieces of paper as a gravitational interaction? please explain wh

y you think that your answer is true.
Physics
1 answer:
mina [271]2 years ago
5 0

It is not possible  to explain the interaction of the rod and pieces of paper as a gravitational interaction.

<h3>What is Gravitational interaction?</h3>

This is defined as the interaction between a particle or body resulting from their mass. This type of interaction is usually weak and occurs in all distances possible.

It is not gravitational interaction, because the rod attracts paper only against the gravitational force of the earth and there is no attraction between both bodies under a different condition.

This is therefore the reason why it is not possible to explain the interaction of the rod and pieces of paper as a gravitational interaction.

Read more about Gravitational interaction here brainly.com/question/25624188

#SPJ1

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Apollo astronauts took a "nine iron" to the Moon and hit a golf ball about 180 m.
Studentka2010 [4]
So in calculating this one its is really hard to explain how i get it on solve it but you must consider this factors that i give in getting the answer. First is the distance cover by the ball when it is hit by the club, Second is you must estimate both of those data when it is in the moon and in the earth whre the gravity of the earth is 9.8m/s^2 so by calculating the Gravity of the moon or gMoon is equal 1.74m/s^2
7 0
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A 0.09-kg lead bullet traveling 182 m/s strikes an armor plate and comes to a stop. If all of the bullet's energy is converted t
Artemon [7]

Answer: Δθ = 127.4 K

Explanation: by using the law of conservation of energy, the kinetic energy of the bullet equals the heat energy on the plate.

Kinetic energy of bullet = mv²/2

Heat energy = mcΔθ

Where m = mass of bullet = 0.09kg, v = velocity of bullet = 182 m/s, c = specific heat capacity of lead bullet = 130 j/kgk

Δθ = change in temperature

mv²/2 = mcΔθ

With 'm' on both sides of the equation, they cancel out each other, hence we have that

v²/2 = cΔθ

v² = 2cΔθ

Δθ= v²/2c

Δθ = (182)²/2×130

Δθ = 33124/260

Δθ = 127.4 K

3 0
3 years ago
Why do solids have a definite shape?
ddd [48]
Any matter that is a solid<span> has a </span>definite shape<span> and a </span>definite<span> volume. The molecules in a </span>solid<span> are in fixed positions and are close together. Although the molecules can still vibrate, they cannot move from one part of the </span>solid<span> to another part. As a result, a </span>solid does<span> not easily change its </span>shape<span> or its volume</span>
5 0
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Calculate the force of gravity between planet X and planet y if both planets are 3.75 X 10^11 m apart, planet X has a mass of 1.
GenaCL600 [577]

So, the force of gravity that the asteroid and the planet have on each other approximately \boxed{\sf{2.9 \times 10^{17} \: N}}

<h3>Introduction</h3>

Hi ! Now, I will help to discuss about the gravitational force between two objects. The force of gravity is not affected by the radius of an object, but radius between two object. Moreover, if the object is a planet, the radius of the planet is only to calculate the "gravitational acceleration" on the planet itself,does not determine the gravitational force between the two planets. For the gravitational force between two objects, it can be calculated using the following formula :

\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}

With the following condition :

  • F = gravitational force (N)
  • G = gravity constant ≈ \sf{6.67 \times 10^{-11}} N.m²/kg²
  • \sf{m_1} = mass of the first object (kg)
  • \sf{m_2} = mass of the second object (kg)
  • r = distance between two objects (m)

<h3>Problem Solving</h3>

We know that :

  • G = gravity constant ≈ \sf{6.67 \times 10^{-11}} N.m²/kg²
  • \sf{m_X} = mass of the planet X = \sf{1.55 \times 10^{22}} kg.
  • \sf{m_Y} = mass of the planet Y = \sf{3.95 \times 10^{28}} kg.
  • r = distance between two objects = \sf{3.75 \times 10^{11}} m.

What was asked :

  • F = gravitational force = ... N

Step by step :

\sf{F = G \times \frac{m_X \times m_Y}{r^2}}

\sf{F = 6.67 \cdot 10^{-11} \times \frac{1.55 \cdot 10^{22} \cdot 3.95 \times 10^{28}}{(3.75 \times 10^{11})^2}}

\sf{F \approx \frac{40.84 \times 10^{-11 + 22 + 28}}{14.0625 \times 10^{22}}}

\sf{F \approx 2.9 \times 10^{39 - 22}}

\sf{F \approx 2.9 \times 10^{17} \: N}

<h3>Conclusion</h3>

So, the force of gravity that the asteroid and the planet have on each other approximately

\boxed{\sf{2.9 \times 10^{17} \: N}}

<h3>See More</h3>
  • Gravity is a thing has depends on ... brainly.com/question/26485200
8 0
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What was the voltage on the battery for the #2 series circuit with two light bulbs?A: 4 VB: 18 VC: 4.5 VD: 9 V
Ghella [55]

We are given a series circuit with two light bulbs. In this case, the light bulbs act as resistors in series and the total resistance is:

R_t=R_1+R_2

That is the sum of all the resistances in series in the circuit. To determine the voltage we can use Ohm's law:

V=IR

Where "R" is the total resistance and "I" is the current in the circuit. Replacing we get:

V=I(R_1+R_2)

8 0
1 year ago
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