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bonufazy [111]
3 years ago
13

Two forces are going in opposite directions each force is 9.

Physics
1 answer:
professor190 [17]3 years ago
7 0

Answer:

The net force is zero.

Explanation:

Two opposing and equal forces cancel each other out, giving you a net force of zero.

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A soccer player applies a force of 48.4 N to a soccer ball while kicking it. If the ball has
AlekseyPX

Answer:

C. 110 m/s2

Explanation:

Force = Mass x Acceleration

Since we have the force and the mass, we can rearrange this equation to solve for acceleration by dividing both sides by mass:

Force/Mass = (Mass x Acceleration)/Mass

Acceleration = Force/Mass

Now we just have to plug in our values and calculate!

Acceleration = 48.4/0.44

Acceleration = 110m/s/s

It is option C. 110 m/s2

Hope this helped!

6 0
3 years ago
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How much tension must a rope withstand if used to accelerate a 960-kg car from rest horizontally along a frictionless surface to
Svetach [21]
Use a=(dv/dt)         (change in velocity/ change in time)=acceleration
(1.2/5)=acceleration

F=ma (Newton's second law, Force= Mass x Acceleration

=960 x 0.24     F=230.4N If  T<230.4N then the tow rope will hold

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3 years ago
What living things began to increase in numbers following this mass extinction?
RoseWind [281]

pigsExplanation: population

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3 years ago
If an piano and an apple are dropped in a tunnel without air.which would hit the ground first
liq [111]

Answer:

the piano.

Explanation:

4 0
3 years ago
the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
3 years ago
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