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timama [110]
4 years ago
13

If an object were released in space far away from planets or stars and given an initial momentum (mv), describe what would happe

n to that object.
Physics
2 answers:
Jobisdone [24]4 years ago
5 0
This is a perfect time to review Newton's first law of motion:

"A body remains in uniform motion until acted on by an external force."

Uniform motion means zero acceleration, and THAT means
constant speed in a straight line.

The way you described your object, there's no external force
acting on it.  So it continues in uniform motion ... straight line,
constant speed, constant momentum, zero acceleration.
erik [133]4 years ago
4 0
Would just keep going with the initial momentum? There is no air resistance and if it is far enough away from planets or stars than gravity really would affect it so it couldn't be pulled in to a planet/star or even orbit them. I'm pretty sure that's what would happen. 
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A ball is thrown vertically upward with a speed of 27.9 m/s from a height of 2.0 m. How long does it take to reach its highest p
Bas_tet [7]

Answer:

2.84403 seconds

2.91483 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-27.9}{-9.81}\\\Rightarrow t=2.84403\ s

It takes 2.84403 seconds to reach the highest point

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-27.9^2}{2\times -9.81}\\\Rightarrow s=39.67431 m

The ball will travel 39.67431+2 = 41.67431 m while going down to the ground

s=ut+\frac{1}{2}at^2\\\Rightarrow 41.71479=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{41.67431\times 2}{9.81}}\\\Rightarrow t=2.91483\ s

The ball takes 2.91483 seconds to hit the ground after it reaches its highest point.

6 0
3 years ago
F all of the energy in a falling object's gravitational potential energy store is transferred to its kinetic energy store by the
stepladder [879]

Answer:

The options are not shown, so let's derive the relationship.

For an object that is at a height H above the ground, and is not moving, the potential energy will be:

U = m*g*H

where m is the mass of the object, and g is the gravitational acceleration.

Now, the kinetic energy of an object can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Now, when we drop the object, the potential energy begins to transform into kinetic energy, and by the conservation of the energy, by the moment that H is equal to zero (So the potential energy is zero) all the initial potential energy must now be converted into kinetic energy.

Uinitial = Kfinal.

m*g*H = (1/2)*m*v^2

v^2 = 2*g*H

v = √(2*g*H)

So we expressed the final velocity (the velocity at which the object impacts the ground) in terms of the height, H.

5 0
3 years ago
386.5<br>- 129.18 what is it​
algol [13]

Answer:

257.32

Explanation:

I jus worked it out on paper. Brainliest please?

5 0
4 years ago
Read 2 more answers
PLEASE ANSWER FAST!!
nataly862011 [7]
I think that it’s false I might be wrong but I want the points
7 0
3 years ago
Read 2 more answers
A wave oscillates 5.0 times a second and has a speed of 4.0m/s what is the frequency of this wave
Viefleur [7K]

Answer:

The frequency of the wave is 5.0Hz

Explanation:

The frequency of a wave is the number of oscillations or revolutions made in a second.

7 0
3 years ago
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