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Ket [755]
4 years ago
7

Which statements describe an object in motion that has no external force acting on it? Check all that apply.

Physics
2 answers:
zzz [600]4 years ago
7 0

It moves at a constant speed.

It moves in the same direction.

Explanation:

As stated by Newton's first law:

"an object in motion which has no external force acting on it will continue its motion at constant speed or it will stay at rest if it was at rest".

This is also related to Newton's second law, which states that the acceleration of an object is proportional to the net force applied to it: therefore, no external force means no acceleration, so no change in the motion of the object.

Therefore, only these two options:

It moves at a constant speed.

It moves in the same direction.

describes objects in motion that have no external force acting on it. On the contrary, all the other options involve a change in the velocity (either speed of direction) of the object, so they describe objects which are accelerating.

svetoff [14.1K]4 years ago
5 0
<span>It moves at a constant speed
</span><span>It moves in the same direction.</span>
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The acceleration of gravity on Moon B is D) 0.25 m/s2

Explanation:

The data listed in the problem are not clear: find them in the table attached.

The acceleration of gravity on a planet is given by the following equation:

g=\frac{GM}{R^2}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

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M=1.5\cdot 10^{20} kg (mass)

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g=\frac{(6.67\cdot 10^{-11})(1.5\cdot 10^{20})}{(2.0\cdot 10^5)^2}=0.25 m/s^2

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3 years ago
Explain why the roller coaster’s potential energy is greater at point 1 than at point 4.
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3 years ago
What is the work done by a car's braking system when it slows the 1500-kg car from an initial speed of 96 km/h down to 56 km/h i
kondor19780726 [428]

Answer:

-352.275KJ

Explanation:

We are given that

Mass of car=1500kg

Initial speed of car =u=96 km/h=96\times \frac{5}{18}=26.67m/s

1km/h=\frac{5}{18}m/s

Final speed of car=v=56km/h=56\times \frac{5}{18}=15.56m/s

Distance traveled by car=s=55m

We have to find the work done by the car's braking system.

Using third equation of motion

v^2-u^2=2as

(15.56)^2-(26.67)^2=2a(55)

-469.18=110a

a=\frac{-469.18}{110}=-4.27m/s^2

Where negative sign indicates that velocity of car decreases.

Work done by a car's barking system=w=F\times s=ma\times s

Work done by a car's barking system=1500\times -4.27\times 55=352275J

Work done by a car's barking system=-\frac{352275}{1000}=-352.275KJ

1KJ=1000J

Where negative sign indicates that work done in opposite direction of motion.

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4 years ago
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