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Ronch [10]
3 years ago
9

What happens to the motion of an object when the forces acting on it are balanced? A. The motion changes. B. The motion does not

change. C. The motion speeds up. D. The motion slows down.
Physics
2 answers:
dexar [7]3 years ago
8 0

D. The motion slows down. Because If they are two forces putting force on the opposite sides, then the object would start to balance out and slow down

S_A_V [24]3 years ago
4 0

Question: What happens to the motion of an object when the forces acting on it are balanced?

Choices:

A. The motion changes.

B. The motion does not change.

C. The motion speeds up.

D. The motion slows down.

Answer: <u>D) The motion slows down when a motion of an object forces acting on the balance</u>.

<em>Hope this helps!.</em>

<em>~~~~~~~~~~~</em>

<em>~A.W~ZoomZoom44</em>

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A person run 21.0 km west then runs 10.0 km east what is the persons distance
VashaNatasha [74]

Answer:

31.0 km

Explanation:

21.0+10.0=31.0 km

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I think its a correct answer

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baseball is hit into the air at an initial speed of 37.2 m/s and an angle of 49.3 ° above the horizontal. At the same time, the
Agata [3.3K]

Answer:

The average speed of the fielder is 5.24 m/s

Explanation:

The position vector of the ball after it was hit can be calculated using the following equation:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem.

When the ball is caught, its position vector will be (see r1 in the figure):

r1 = (r1x, 0.873 m)

Then, using the equation of the position vector written above:

r1x = x0 + v0 · t · cos α

0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²

Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:

r1x = v0 · t · cos α

0.873 m = v0 · t · sin α + 1/2 · g · t²

Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:

0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²

0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²

Solving the quadratic equation:

t = 0.03 s and t = 5.72 s.

It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.

With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:

r1x = v0 · t · cos α

r1x = 37.2 m/s · 5.72 s · cos 49.3°

r1x = 1.39 × 10² m

The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.

The average velocity is calculated as the traveled distance over time, then:

average velocity = treveled distance / elapsed time

average velocity = 30.0 m / 5.72 s = 5.24 m/s

8 0
3 years ago
What property is used to distinguish the layers of the atmosphere?
Margarita [4]
There are many porperties. You can use Altitude, Temperature, Pressure and Density, but the best one is temperature. The resaon for that is that based on the temperature changes then the athmosphere can be broken into four major layers. Remember that the layers are the following: <span>the </span>troposphere,the<span> </span>stratosphere, <span>the </span>mesosphere<span>, and the</span>thermosphere<span>.</span>
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zvonat [6]

I'm going to assume that this gripping drama takes place on planet Earth, where the acceleration of gravity is 9.8 m/s².  The solutions would be completely different if the same scenario were to play out in other places.

A ball is thrown upward with a speed of 40 m/s.  Gravity decreases its upward speed (increases its downward speed) by 9.8 m/s every second.

So, the ball reaches its highest point after (40 m/s)/(9.8 m/s²) = <em>4.08 seconds</em>. At that point, it runs out of upward gas, and begins falling.

Just like so many other aspects of life, the downward fall is an exact "mirror image" of the upward trip.  After another 4.08 seconds, the ball has returned to the height of the hand which flung it.  In total, the ball is in the air for <em>8.16 seconds</em> up and down.

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