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2 years ago

Two cars each have a mass of 1050 kg. If the gravitational force between

Physics

2 answers:

Cloud [144]2 years ago

7 0

Answer: D.

Explanation:

d = ± √ G m 1 m 2 /F

Because distance cannot be negative there it goes:

d = √ G m 1 m 2 /F

d = √ 6.67 ⋅ 10 − 11 ⋅ 1050 ⋅ 1050 2.27 ⋅ 10 − 7

d = √ 0.0000735 2.27 ⋅ 10 − 7

d = √ 3239504.405

d ≈ 1799.86 m = 18m

guapka [62]2 years ago

7 0

Answer:

18 m

Explanation:

Aye pex... its very painful

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What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each

Bezzdna [24]

This question apparently comes after an EARLIER one,
where you were told either the voltage across the same
capacitor or the total charge stored in it. You can't answer
THIS one without that information.

3 0

3 years ago

Answer the question below please I give brainliest

Basile [38]

Answer: The first one

Explanation: I think it's the first one because it says what is the "least" gravitational potential energy story between the prairie dog and Earth that said resting in its borrow is using less energy

7 0

2 years ago

Read 2 more answers

the force that is exerted when a shopping cart is pushed. the forces that causes a metal ball to move toward a magnet

alexandr402 [8]

Answer:

The force that is exerted when a shopping cart is pushed is a type of push force, supplied by the muscles of the cart pusher's body.

The forces that causes a metal ball to move toward a magnet is a type of pull force that is as a result of the magnetic field forces.

Explanation:

Forces are divided into push forces that tends to accelerate a body away from the source of the force, and pull forces that accelerates the body towards the force source.

Examples of push forces includes pushing a cart, pushing a table, repulsion of two similar poles of a magnet etc. Examples of pull forces includes a attractive force between two dissimilar poles of a magnet, pulling a load by a rope, a dog pulling on a leash etc.

8 0

2 years ago

Read 2 more answers

Can you explain that gravity pulls us to the Earth & can you calculate weight from masses on both on Earth and other planets

schepotkina [342]

I don't actually understand what your question is, but I'll dance around the subject
for a while, and hope that you get something out of it.

-- The effect of gravity is: There's a pair of forces, in both directions, between
every two masses.

-- The strength of the force depends on the product of the masses, so it doesn't matter whether there's a big one and a small one, or whether they're nearly equal.
It's the product that counts. Bigger product ==> stronger force, in direct proportion.

-- The strength of the forces also depends on the distance between the objects' centers. More distance => weaker force. Actually, (more distance)² ==> weaker force.

-- The forces are equal in both directions. Your weight on Earth is exactly equal to
the Earth's weight on you. You can prove that. Turn your bathroom scale face down
and stand on it. Now it's measuring the force that attracts the Earth toward you.
If you put a little mirror down under the numbers, you'll see that it's the same as
the force that attracts you toward the Earth when the scale is right-side-up.

-- When you (or a ball) are up on the roof and step off, the force of gravity that pulls
you (or the ball) toward the Earth causes you (or the ball) to accelerate (fall) toward the Earth.
Also, the force that attracts the Earth toward you (or the ball) causes the Earth to accelerate (fall) toward you (or the ball).
The forces are equal. But since the Earth has more mass than you have, you accelerate toward the Earth faster than the Earth accelerates toward you.

-- This works exactly the same for every pair of masses in the universe. Gravity
is everywhere. You can't turn it off, and you can't shield anything from it.

-- Sometimes you'll hear about some mysterious way to "defy gravity". It's not possible to 'defy' gravity, but since we know that it's there, we can work with it.
If we want to move something in the opposite direction from where gravity is pulling it, all we need to do is provide a force in that direction that's stronger than the force of gravity.
I know that sounds complicated, so here are a few examples of how we do it:
-- use arm-muscle force to pick a book UP off the table
-- use leg-muscle force to move your whole body UP the stairs
-- use buoyant force to LIFT a helium balloon or a hot-air balloon
-- use the force of air resistance to LIFT an airplane.

-- The weight of 1 kilogram of mass on or near the Earth is 9.8 newtons. (That's
about 2.205 pounds). The same kilogram of mass has different weights on other planets. Wherever it is, we only know one of the masses ... the kilogram. In order
to figure out what it weighs there, we need to know the mass of the planet, and
the distance between the kilogram and the center of the planet.

I hope I told you something that you were actually looking for.

7 0

3 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago