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zvonat [6]
4 years ago
13

As you found in Part A, your weight will be greater than normal when the elevator is moving upward with increasing speed. For wh

at other motion would your weight also be greater than your normal weight?
A) The elevator moves upward with constant velocity.
B) The elevator moves downward with constant velocity.
C) The elevator moves upward while slowing in speed.
D) The elevator moves downward while slowing in speed.
E) The elevator moves downward while increasing in speed.
Physics
1 answer:
shepuryov [24]4 years ago
4 0

Answer:

D) The elevator moves downward while slowing in speed.

Explanation:

The elevator is moving downward with decreasing speed , that means a force is acting on it in upward direction . So it has  an acceleration in upward direction . This type of movement is similar to motion with acceleration in upward direction.

In this case apparent weight will be greater.

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A 25,000-kg train car moving at 2.50 m/s collides with and connects to a train car of equal mass moving in the same direction at
Margarita [4]

Answer:

14062.5 J

Explanation:

From the law of conservation of momentum,

Total momentum before collision  = Total momentum after collision.

V = (m₁u₁ + m₂u₂)/(m₁+m₂).................1

Where V = common velocity after collision

Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s

Substitute into equation 1

V = [25000(2.5) + 25000(1)]/(25000+25000)

V = (62500+25000)/50000

V = 87500/50000

V = 1.75 m/s.

Note: The collision is an inelastic collision as such there is lost in kinetic energy of the system.

Total Kinetic energy before collision = kinetic energy of the first train car + kinetic energy of the second train car

E₁ = 1/2m₁u₁² + 1/2m₂u₂²........................ Equation 2

Where E₁ = Total kinetic energy of the body before collision, m₁ and m₂ = mass of the first train car and second train car respectively. u₁ and u₂ = initial velocity of the first train car and second train car respectively.

Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s

Substitute into equation 2

E₁ = 1/2(25000)(2.5)² + 1/2(25000)(1.0)²

E₁ = 12500(6.25) + 12500

E₁ = 78125+12500

E₁ = 90625 J.

Also

E₂ = 1/2V²(m₁+m₂)....................... Equation 3

Where E₂ = total kinetic energy of the system after collision, V = common velocity, m₁ and m₂ = mass of the first and second train car respectively.

Given: V = 1.75 m/s, m₁ = m₂ = 25000 kg

Substitute into equation 3

E₂ = 1/2(1.75)²(25000+25000)

E₂ = 1/2(3.0625)(50000)

E₂ = (3.0625)(25000)

E₂ = 76562.5 J.

Lost in kinetic Energy of the system = E₁ - E₂ = 90625 - 76562.5

Lost in kinetic energy of the system = 14062.5 J

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