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3 years ago

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Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters

per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
PLZ HURRY WILL MARK BRAINLIEST IF CORRECT

2 answers:

vovangra [49]3 years ago

8 0

Answer:

$\boxed{\mathrm{A. \: It \: will \: be \: 0 \: meters \: per \: second \: per \: second. }}$

Explanation:

$\mathrm{force=mass \times acceleration}$

The force is given 0 newtons.

$\mathrm{force=0 \: N}$

Plug force as 0.

$\mathrm{0=mass \times acceleration}$

Divide both sides by mass.

$\mathrm{\frac{0}{mass} =acceleration}$

$\mathrm{0 =acceleration}$

$\mathrm{acceleration= 0\: m/s/s}$

Vanyuwa [196]3 years ago

3 0

Answer:

Option A

Explanation:

Acceleration will be obviously zero when Force = 0

<u><em>That is how</em></u>:

Force = Mass * Acceleration

So, If force = 0

0 = Mass * Acceleration.

Dividing both sides by Mass

Acceleration = 0/Mass

Acceleration = 0 m/s²

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The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

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Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u> $m_{1}$ <u>:</u>

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

<u>For </u> $m_{2}$ <u>:</u>

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

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<u>For </u> $m_{3}$ <u>:</u>

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$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

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