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Klio2033 [76]
3 years ago
8

A box measures 4 m x 3 m x 2 m would it be easier to push the box if it were resting on a 3 m x 2 m side or on the 4 m x 3 m sid

e
Physics
1 answer:
ozzi3 years ago
4 0

Answer:

465408754m468 iui m456m

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Your friend claims that weather predictions are just guesses. What would you say to change your friend's mind?
VashaNatasha [74]
To change my friends mind, after they've already said weather predictions are just guesses, I would say they are correct, but not only to they use guesses to predict what will happen, but they also, before hand, use atmospheric tools. They then predict from the data collected. They do both.
4 0
3 years ago
Which of the following has potential but not kinetic energy?
gavmur [86]

The answer would be option D "a ball sitting on a shelf." Potential energy is the amount of energy a object has while it's at rest.. (or not moving) Kinetic energy is how much energy a object is while it's moving. So in this case it's option D because a ball sitting on a shelf isn't moving therefore it has potential energy. It's not option A because thats a example of kinetic energy since how the roller coaster is moving. It's not option B because it's kinetic energy because the bike is moving. It's also not option C because it's kinetic energy because the bird is moving.


Hope this helps!

7 0
3 years ago
Which element could be described by the list of properties above?
Murljashka [212]

Answer:  

list the properties

Explanation:

7 0
3 years ago
Read 2 more answers
This is physics class, plzzz help ! 86.72 m to cm
Vaselesa [24]

Answer:

8672

Explanation:

multiply the length value by 100

5 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
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