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djyliett [7]
3 years ago
10

The total mechanical energy of a system is equally divided between kinetic energy and potential energy. can never be negative. i

s constant only if conservative forces act. is either all kinetic energy or all potential energy, at any one instant. is constant only if nonconservative forces act. A skier, of mass 60. kg, pushes off the top of a frictionless hill with an initial speed of 4.0 m/s. How fast will she be moving after dropping 10. m in elevation? 15. m/s 0.15 km/s 0.20 km/s 49. m/s 10. m/s In an inelastic collision, the final total momentum is less than the initial momentum. the same as the initial momentum. insufficient information to answer more than the initial momentum. A handball of mass 0.10 kg, traveling horizontally at 25. m/s, strikes a wall and rebounds at 19. m/s. What is the change in the momentum of the ball? 1.2 N-s 72 N-s 5.4 N-S 1.8N-s 4.4 N-s Car A (mass = 1000 kg) moves to the right along a level, straight road at a speed of 6.0 m/s. It collides directly with car B (mass = 200 kg) in a completely inelastic collision. What is the momentum after the collision if car B was initially at rest? 10. KN-s to the right zero 6.0 KN-s to the right 8.0 KN-s to the left 2.0 KN-s to the right
Physics
1 answer:
inn [45]3 years ago
8 0

Answer:

1-  is constant only if conservative forces act

2- 60*9.8*10+0.5*60*4^2=0.5*60*v^2

v=14.56=15 m/s

3- B (Momentum is conserved)

4- Change in momentum= m(v-u) = 0.10(19--25)=4.4 Ns

5- 1000*6+0 = 6000 KN Momentum before to the right, So final momentum should also be same.

Explanation:

5 questions are in 1 question all of them are answered.

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In an experiment, an object is released from rest from the top of a building. Its speed is measured as it reaches a point that i
NikAS [45]

Answer:

so the speed will increase by 1.44 times then the initial speed if the distance is increased to double

Explanation:

As we know that the air friction or resistance due to air is neglected then we can use the equation of kinematics here

v_f^2 - v_i^2 = 2 a d

since we released it from rest so we have

v_i = 0

so here we have

v_f = \sqrt{2gd}

now if the distance is double then we have

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now from above two equations we can say that

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so the speed will increase by 1.44 times then the initial speed if the distance is increased to double

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Explanation:

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The following graph shows the force exerted on and the displacement of object being pulled
Tomtit [17]

The work done to pull the object 7.0 m is the total area under the graph from 0.0 m to 7.0 m, determined as 245 J.

<h3>Work done by the applied force</h3>

The area under force versus displacement graph is work done.

The total work done by pulling the object 7 m, can be grouped into two areas;

  • First area, A1 = area of triangle from 0 m to 2.0 m
  • Second area, A2 = area of trapezium, from 2.0 m to 7.0 m

A1 = ¹/₂ bh

A1 = ¹/₂ x (2) x (20)

A1 = 20 J

A2 = ¹/₂(large base + small base) x height

A2  = ¹/₂[(7 - 2) + (7-3)] x 50

A2 = ¹/₂(5 + 4) x 50

A2 = 225 J

<h3>Total work done </h3>

W = A1 + A2

W = 20 J + 225 J

W = 245 J

Learn more about work done here: brainly.com/question/8119756

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<em>r=x+y</em>

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