What are the two factors that affect the gravitational pull between two objects?

Physics

1 answer:

elena-14-01-66 [18.8K]3 years ago

4 0

Answer:

hen dealing with the force of gravity between two objects, there are only two things that are important – mass, and distance.

Explanation:

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Master of physics needed

Delicious77 [7]

Hey JayDilla, I get 1/3. Here's how:
Kinetic energy due to linear motion is:
$E_{linear}= \frac{1}{2}mv^2$
where
$v=r \omega$
giving
$E_{linear}= \frac{1}{2}mr^2 \omega ^2$

The rotational part requires the moment of inertia of a solid cylinder
$I_{cyl} = \frac{1}{2}mr^2$
Then the rotational kinetic energy is
$E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2$
Adding the two types of energy and factoring out common terms gives
$\frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2})$
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part. Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.

8 0

4 years ago

A child pushes a 75 N toy car across the floor. What is the mass of the car?

GaryK [48]

Answer:

7.6 kg

Explanation:

w=75N

w=mg

m=w÷g

m=75÷9.8

m=7.6kg

8 0

3 years ago

In a rigid container, as the temperature of a gas decreases, the pressure of the gas will decrease. t f

Sergio039 [100]

The answer is true: the pressure of a gas will decrease as temperature decreases in a rigid container.

This is one of the central gas laws called the Gay-Lussac law that states for a given gas at a constant volume, the pressure of the gas is directly proportional to its temperature. We also know that as temperature reduces, so too does molecular interaction. Increased temperature results in increased pressure, and decreased temperature therefore results in decreased pressure.

6 0

4 years ago

Read 2 more answers

Find the moment of inertia Ihoop of a hoop of radius r and mass m with respect to an axis perpendicular to the hoop and passing

Juliette [100K]

Answer: MR²

is the the moment of inertia of a hoop of radius R and mass M with respect to an axis perpendicular to the hoop and passing through its center

Explanation:

Since in the hoop , all mass elements are situated at the same distance from the centre , the following expression for the moment of inertia can be written as follows.

I = ∫ r² dm

= R²∫ dm

MR²

where M is total mass and R is radius of the hoop .

3 0

3 years ago

The collision between a hammer and a nail can be considered to be approximately elastic. estimate the kinetic energy acquired by

Setler [38]

Here we can use momentum conservation as in this type of collision there is no external force on it

$m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}$