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3 years ago

If the distance between the center of two objects is quadrupled. The gravitational

Physics

1 answer:

Juliette [100K]3 years ago

6 0

Answer:

F' = F/16

Explanation:

The gravitational force between masses is given by :

$F=G\dfrac{m_1m_2}{r^2}$

If the distance between the center of two objects is quadrupled, r' = 4r

New force will be :

$F'=G\dfrac{m_1m_2}{r'^2}\\\\F'=G\dfrac{m_1m_2}{(4r)^2}\\\\F'=\dfrac{Gm_1m_2}{16r^2}\\\\F'=\dfrac{1}{16}\times \dfrac{Gm_1m_2}{r^2}\\\\F'=\dfrac{F}{16}$

So, the new force will change by a factor of 16.

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Which describes the position on a horizontal number line

Savatey [412]

Answer:

It is formed by a horizontal number line, called the x-axis, and a vertical number line, called the y-axis.

Explanation:

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3 years ago

How can we describe, model, and explain the material world, and are there limitations?

taurus [48]

Answer:· A model is a description of natural phenomenon that scientists can use to make predictions. A good model is both as accurate as possible and as simple as possible, which makes it not only powerful but also easy to understand. However, no matter how good they are, models will almost always have limitations.

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3 years ago

What is the main fuel consumed in the core of a red giant? a. H b. C c. Fe d. He

vladimir1956 [14]

What is the main fuel consumed in the core of a red giant?
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3 years ago

What is the distance from axis about which a uniform, balsa-wood sphere will have the same moment of inertia as does a thin-wall

andrey2020 [161]

Answer:

$D_{s}$ ≈ 2.1 R

Explanation:

The moment of inertia of the bodies can be calculated by the equation

I = ∫ r² dm

For bodies with symmetry this tabulated, the moment of inertia of the center of mass

Sphere $Is_{cm}$ = 2/5 M R²

Spherical shell $Ic_{cm}$ = 2/3 M R²

The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass

I = $I_{cm}$ + M D²

Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation

Let's start with the spherical shell, axis is along a diameter

D = 2R

Ic = $Ic_{cm}$ + M D²

Ic = 2/3 MR² + M (2R)²

Ic = M R² (2/3 + 4)

Ic = 14/3 M R²

The sphere

Is = $Is_{cm}$ + M [ $D_{s}$ ²

Is = Ic

2/5 MR² + M $D_{s}$ ² = 14/3 MR²

$D_{s}$ ² = R² (14/3 - 2/5)

$D_{s}$ = √ (R² (64/15)

$D_{s}$ = 2,066 R

3 0

3 years ago

Sayid made a chart listing data of two colliding objects. A 5-column table titled Collision: Two Objects Stick Together with 2 r

Alborosie

Answer:

6 m/s is the missing final velocity

Explanation:

From the data table we extract that there were two objects (X and Y) that underwent an inelastic collision, moving together after the collision as a new object with mass equal the addition of the two original masses, and a new velocity which is the unknown in the problem).

Object X had a mass of 300 kg, while object Y had a mass of 100 kg.

Object's X initial velocity was positive (let's imagine it on a horizontal axis pointing to the right) of 10 m/s. Object Y had a negative velocity (imagine it as pointing to the left on the horizontal axis) of -6 m/s.

We can solve for the unknown, using conservation of momentum in the collision: Initial total momentum = Final total momentum (where momentum is defined as the product of the mass of the object times its velocity.

In numbers, and calling $P_{xi}$ the initial momentum of object X and $P_{yi}$ the initial momentum of object Y, we can derive the total initial momentum of the system: $P_{total}_i=P_{xi}+P_{yi}= 300*10 \frac{kg*m}{s} -100*6\frac{kg*m}{s} =\\=(3000-600 )\frac{kg*m}{s} =2400 \frac{kg*m}{s}$

Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):

Final momentum of the system: $M * v_f=400kg * v_f$

We then set the equality of the momenta (total initial equals final) and proceed to solve the equation for the unknown(final velocity of the system):

$2400 \frac{kg*m}{s} =400kg*v_f\\\frac{2400}{400} \frac{m}{s} =v_f\\v_f=6 \frac{m}{s}$

7 0

4 years ago

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