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MrRa [10]
3 years ago
10

Frank does 2400J of work in climbing a set of stairs. If he does the work in 6 seconds, what is his power output?

Physics
1 answer:
timurjin [86]3 years ago
4 0
Power = Work Done / Time taken

         Where Work Done is Joules, and Time is in Seconds, Power is in Watts

            = 2400 J / 6 seconds

            = 400 Watts

The power output is 400 Watts.
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The ice skaters partner liftes her up a distance of 1 m work done or not work done
SOVA2 [1]

Answer:

Work done.

Explanation:

The skater who lifts has to overcome the partner's weight. When lifted up by 1 meter, her potential energy increases by (mass)x(gravitational acceleration)x(1meter), which is the amount of work done.

(This all assumes lifting vertically and no other forces being part of the picture)

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What two major uses does this (H-R)diagram have for astronomers?
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7 0
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What is the formula you use to determine the gravitational potential energy of a object
kozerog [31]

Gravitational potential energy, relative to some level =

       (mass of the object)
times
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4 0
3 years ago
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

3 0
3 years ago
A car was moving at 14 m/s After 30 s, its speed increased to 20 m/s. What was the acceleration during this time ( need help fas
Arada [10]

Answer:let initial velocity u=14m/s

Final velocity v=20m/s

Time taken t=30

Acceleration =a

V=u +at

a= (20-14)/30

a=0.2m/s^2

Explanation:

Acceleration is the change in velocity with respect to time.

7 0
3 years ago
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