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Mashcka [7]
3 years ago
10

Objects 1 and 2 attract each other with a gravitational force of 45 units. If the mass of Object 1 is doubled, then the new grav

itational fore will be ______ units.
Physics
2 answers:
Novosadov [1.4K]3 years ago
4 0

Explanation:

Fgravity = G*(mass1*mass2)/D²

so, if you double one of the masses, what does that do to our equation ?

Fgravitynew = G*(2*mass1*mass2)/D²

due to the commutative property of multiplication

Fgravitynew = 2* G*(mass1*mass2)/D² = 2* Fgravity

so, the correct answer will be 2×45 = 90 units.

kirza4 [7]3 years ago
3 0
If the gravitational force is 45 and we double its force, we are only doing 45 multiplied by 2 since double is two. 45 multiplied by 2 is equal to 90.

ANSWER:

If the mass of object 1 is doubled, then the new gravitational force will be 90 units.
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A car slows down at -5.00 m/s² until it comes to a stop after travelling 15.0 m. What was the initial speed of the car?
lapo4ka [179]
<h2>Answer: 12.24m/s</h2>

According to <u>kinematics</u> this situation is described as a uniformly accelerated rectilinear motion. This means the acceleration while the car is in motion is constant.

Now, among the equations related to this type of motion we have the following that relates the velocity with the acceleration and the distance traveled:

V_{f}^{2}-V_{o}^{2}=2.a.d   (1)

Where:

V_{f} is the Final Velocity of the car. We are told "the car comes to a stop after travelling", this means it is 0.

V_{o} is the Initial Velocity, the value we want to find

a is the constant acceleration of the car (the negative sign means the car is decelerating)

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Now, let's substitute the known values in equation (1) and find V_{o}:

0-V_{o}^{2}=2(-5m/{s}^{2})(15m)    (2)

-V_{o}^{2}=-150{m}^{2}/{s}^{2}    (3)

Multiplying by -1 on both sides of the equation:

V_{o}^{2}=150{m}^{2}/{s}^{2}    (4)

V_{o}=\sqrt{150{m}^{2}/{s}^{2}}    (5)

Finally:

V_{o}=12.24m/s >>>This is the Initial velocity of the car

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