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Eva8 [605]
2 years ago
13

An object has a 600 N force pushing it to the right, while a 700 N force pushes in the opposite direction. Gravity also acts on

the object with a force of 400N, while the normal force pushes upward with a force of 400 N.
Which direction will the object move in?


to the right

to the left

downward

upward
Physics
2 answers:
nordsb [41]2 years ago
6 0
Left is the answer 3
love history [14]2 years ago
6 0

Answer:

the ans is left

Explanation:

because the LHS has very les frictin acting on it

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2 years ago
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A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.820 m. W
Vikentia [17]

Answer:

1.0752 kgm/s

Explanation:

Considering when the drop was dropped from rest from a height,

mass of the ball, m = 0.120 kg

height, h = - 1.25 m

the initial velocity, u = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            V^{2} = U^{2} + 2gh

Substituting the values,

                             V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)

                             V^{2} = 24.5 m/s

                             V = \sqrt{24.5} \ m/s

                             V = 4.95 \ m/s

                            V = ± 4.95 m/s

                            V = - 4.95 m/s

Since the ball is moving downward, the final velocity of the ball when it hits the floor is  V = - 4.95 m/s  

Considering when the ball rebounds from the floor,

assume the mass of the ball still remain, m = 0.120 kg

height, h = 0.820 m

the final velocity, v = 0 m/s  

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            V^{2} = U^{2} + 2gh

Substituting the values,

                            0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)

                            0 = U^{2} - 16.072 m/s

                            U^{2} = 16.072 m/s

                            U = \sqrt{16.072} \ m/s

                           U = ± 4.01 m/s

                          U = + 4.01 m/s

Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is  U = + 4.01 m/s                        

From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.

                            F = \frac{mv - mu}{t}

                          F.t = m(v - u)

       ⇒      Impulse = Change in momentum

To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,                        

          ∴          F.t = 0.120  kg(4.01  m/s - (-4.95  m/s) )

                      F.t = 0.120  kg(4.01  m/s + 4.95  m/s) )

                      F.t = 0.120  kg × 8.96  m/s

                      Impulse  = 1.0752 kgm/s

The impulse given to the ball by the floor is 1.0752 kgm/s

                             

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