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zzz [600]
2 years ago
12

Which statement is always true about objects which exert gravitational force on each other? A. They are a very large size. B. Th

ey exist only in outer space. C. They are a speed. D.They are a mass.​
Physics
1 answer:
Keith_Richards [23]2 years ago
7 0

Answer:

The answer is D. They are a mass.

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Anne releases a stone from a height of 2 meters. She measures the kinetic energy of the stone at 9.8 joules at the exact point i
insens350 [35]
A. 0.5kg

To get this answer you need to follow the equation of KE=0.5*mv^2 
But we don't have the m part in the equation. So just plug in the numbers to see which works best, though I can tell you before we do that the answer would be a. 

As you may know, gravity, is a force of 9.8 m/s. And we want to get 9.8 Joules. So if we take a half a kg stone, release it at one meter, we get half of the normal gravity pull, 4.90 Joules. That means if we take half a kg stone and drop it at a doubled height, we get 9.8 Joules.

That is also to say that if we have a 1kg stone and drop it at one meter you will get the normal pull of gravity in Joules, 9.8J. 

Be careful though, this does not mean if you drop a 1kg stone and a .5 kg stone the 1kg will hit first. This simply means that the 1kg stone will have twice the Joules that the .5kg stone has.
7 0
3 years ago
Read 2 more answers
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

t=11.2\,\mathrm s

At this time, the rocket attains a velocity of

v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}

When it's in freefall, the rocket's altitude is given by

y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity, and its velocity is

v_2(t)=124\dfrac{\rm m}{\rm s}-gt

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for y_2(t) to reach 0:

1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

{v_f}^2-{v_i}^2=2a\Delta y

where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)

Solve for y_{\rm max} and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to y_2(t)) to be about 32.6 s. Plug this into v_2(t) to find the velocity before it crashes:

v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0
3 years ago
Question 3 of 10
AlexFokin [52]
The answer is Jupiter
6 0
2 years ago
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Calculate the elastic potential energy stored in a spring if it has a force constant of 150 N/m. the spring is extended to a len
Alenkinab [10]

Answer:

6.75J

Explanation:

U=1/2KΔx²

U=0.5* 150*0.30^2

4 0
2 years ago
A geologist is trying to determine what time period a layer of rock was formed.
vodka [1.7K]

Answer:

B

Explanation:

You always want to test as many samples as possible

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2 years ago
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