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ycow [4]
3 years ago
13

A 10.0 kg box is dragged 5.00 meters across a rough horizontal floor by a rope with a tension

Physics
1 answer:
Lemur [1.5K]3 years ago
5 0

Answer:

Explanation:

Net work done on the box = increase in kinetic energy

= 1 / 2 x 10 ( 5² - 3² )

= 80 J .

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mina [271]

Answer:

This link was diagram

Explanation:

https://doubtnut.app.link/FnsNC80Dccb

7 0
3 years ago
PLEASE HELP ASAP
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Aluminum comes from several sources; it rarely occurs in pure form in nature, and is most frequently found embedded in other minerals, primarily bauxite. This would be most commonly found in the dirt.
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Three rocks, each of varying mass and volume, are released from a window at the same time. The window is at height h above the g
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All three rocks will achieve the same speed
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3 years ago
Suppose a car traveling at 8m/s is brought to rest in a distance of 20m.what is it's deceleration and time taken.
LiRa [457]

Answer:

         Time - taken = 2.5 s

          deceleration= -8 m/s²

Solution:

            Given:

                     speed, v = 8 m/s

                 distance, d = 20m

                     

              To Find:

                     deacceleration = ?

               

               As we know speed is defined as

                          v = d/t

                plugging in the values

                          t =  20/ 8

                          t = 2.5s

                Now from deceleration formula

                        a =  - v/ t

                        a = - 20/ 2.5

                        a = - 8 m/s²

          Thus, the time taken and acceleration is 2.5 s and -8 m/s²

          respectively.

          Learn more about deceleration here:

                brainly.com/question/13354629

                       #SPJ4  

               

                       

             

7 0
2 years ago
In the absence of air resistance two balls are thrown upward from the same launch point. Ball a rises to a maximum height above
diamong [38]

Answer:

Va is two times greater than Vb

Explanation:

The maximum height reached by the balls are:

Ymax = \frac{Vo^2}{2g}

Since we are told that Ya = 4Yb:

Ya =  \frac{Va^2}{2g} = 4* Yb = 4 * \frac{Vb^2}{2g}

Simplifying the equation:

Va^2 = 4*Vb^2

Va = 2*Vb

5 0
3 years ago
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