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Ksenya-84 [330]
3 years ago
15

What acceleration results from exerting a 125N force on a 0.65kg ball?

Physics
1 answer:
labwork [276]3 years ago
5 0
first you do your pyramid f is on top and ma is
on bottom were m=mass and a=acceleration
were in this case you do f÷m=force÷mass so again in this case 125÷0.65=192.3 to be more accurate 192.3076923077
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3 years ago
Solve for work when
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So, <u>the value of the work is approximately 84.65 J</u>.

<h2>Introduction</h2>

Hi ! Here I will help you to discuss the subject about work that caused by force in amount value of angle. Work is affected by the force and displacement.

  • If related to the magnitude of the force, the amount of work will be proportional to the magnitude of the applied force. Thats mean, if the value of the force that applied on it is greater, then the value of the work will be greater.
  • If related to the magnitude of shift, the amount of work will be proportional to the magnitude of shift of object. Thats mean, if the value of the shift on it is greater, then the value of the work will be greater.
<h3>Formula Used</h3>

The work done by a moving object can be expressed in the equation:

If the Angle Is Ignored

\boxed{\sf{\bold{W = F \times s}}}

If the Angle Effect on Work

\boxed{\sf{\bold{W = F \times s \times \cos(\theta)}}}

With the following condition:

  • W = work that done by object (J)
  • F = force that applied (N)
  • s = shift or distance (m)
  • \sf{\theta} = angle of elevation (°)

<h3>Solution</h3>

We know that :

  • F = force that applied = \sf{1.41 \times 10^4} N
  • s = shift or distance = 84.9 m
  • \sf{\theta} = angle of elevation = 45°

What was asked ?

  • W = work that done by object = ... J

Step by step :

\sf{W = F \times s \times \cos(\theta)}

\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \cos(45^o)}

\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \frac{\sqrt{2}}{2}}

\sf{W = 119.709 \times \frac{\sqrt{2}}{2}}

\sf{W = 59.8545 \sqrt{2}}

\boxed{\sf{W \approx 84.65 \: J}}

<h3>Conclusion</h3>

So, the value of the work is approximately 84.65 J.

3 0
1 year ago
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