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Wittaler [7]
2 years ago
7

A 70 n force pulls a box straight up 5 m above the ground. how much work is done on the box?

Physics
1 answer:
lord [1]2 years ago
6 0

The work done on the box is 350 joules

Workdone can be described as the amount of energy transferred from an object.

Force can be described as the push on an object, this push makes the object to go through a change in velocity.

Distance can be described as the movement of an object towards any given direction.

It can be calculated by multiplying the force by the distance

Workdone= force × distance

= 70 × 5

= 350

Thus, the work done on the box is 350 joules


Please see the link below for more information
brainly.com/question/14330400?referrer=searchResults


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If a train going 60 m/s hits the brakes, and it takes the train 85 seconds to stop, what is
Anvisha [2.4K]

Question :-

  • If a Train going with 60 m/s Speed hits the Brakes, and takes 85 sec to stop. What is the Acceleration of the Train ?

Answer :-

  • Acceleration of Train is 0.70 m/s² .

Explanation :-

As per the provided information in the given question, we have been given that the Speed of the Train is 60 m/s . Time taken to stop the Train is 85 sec . And, we have been asked to calculate the Acceleration of the Train .

For calculating the Acceleration , we will use the Formula :-

\bigstar \:  \:  \:  \boxed { \sf{ \: Acceleration \:  =  \:  \dfrac{v \:  -  \: u}{t} \:  }} \\

Where ,

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Therefore , by Substituting the given values in the above Formula :-

\dag \:  \:  \:  \:  \sf{Acceleration \:  =  \:  \dfrac{Final  \: Velocity \:  -  \: Initial \: Velocity}{Time} } \\

\longmapsto \:  \:  \: \sf{Acceleration \:  =  \:  \dfrac{0 \:  -  \: 60}{85} } \\

\longmapsto \:  \:  \: \sf{Acceleration \:  =  \:  \dfrac{60}{85} } \\

\longmapsto \:  \:  \: \textbf {\textsf {Acceleration \:  =  \: 0.70 }}

Hence :-

  • Acceleration = 0.70 m/s² .

\underline {\rule {210pt} {4pt}}

Note :-

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Answer:

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Explanation:

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A motorcycle accelerates uniformly from rest, then initial velocity v_i = 0 m/s

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a_T =(V_f - v_i)/t

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Then;  the magnitude of angular acceleration

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