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anzhelika [568]
3 years ago
10

Your friend provides a solution to the following problem. Evaluate her solution. The problem: Jim (mass 50 kg) steps off a ledge

that is 2.0 m above a platform that sits on top of a relaxed spring of force constant 8000 N/m. How far will the spring compress while stopping Jim? Your friend's solution: (50kg)(9.8m/s2)(2.0m)=(1/2)(8000N/m)x x=0.25m4.
Physics
1 answer:
Mekhanik [1.2K]3 years ago
5 0

Answer:

No its wrong

Correct compression is 0.41

Explanation:

After jumping from 2m height on spring attached platform platform is compressed a distance x(let).

So work done by gravity on Jim is converted into spring potential energy.

k=8000 N/m

mass of Jim =50 kg

\frac{1}{2}k(x)^{2} =m*g(2+x)

\frac{1}{2}8000(x)^{2} =50*9.8(2+x)

8.16x^{2} =2 +x

Solve this quadratic one solution is positive and other is negative.

positive one is our answer = 0.41 m

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endothermic reactions

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SOVA2 [1]

Answer:

660 J/kg/°C

Explanation:

Heat lost by metal = heat gained by water

-m₁C₁ΔT₁ = m₂C₂ΔT₂

-(0.45 kg) C₁ (21°C − 80°C) = (0.70 kg) (4200 J/kg/°C) (21°C − 15°C)

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8 0
3 years ago
1. Al descomponerse su vehículo una persona tira de su auto con la ayuda de una cuerda con una fuerza de 3500 N que forma un áng
OleMash [197]

Responder:

Fy = 2474,8737

Fx = 2474,8737

Explicación:

Dado que :

Dado:

Fuerza, F = 3500 N

Ángulo formado con la horizontal, θ, = 45 °

Los componentes de una fuerza se pueden descomponer en componentes verticales y horizontales.

El componente vertical Fy; y

El componente horizontal Fx

Fy = Fuerza * sinθ

Fy = 3500 * sin45 °

Fy = 2474,8737

El componente horizontal:

Fx = Fuerza * cosθ

Fy = 3500 * cos45 °

Fy = 2474,8737

8 0
2 years ago
A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they
Alinara [238K]

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁  +  m₂u₂

P₁ = (1000 x 25)  +  (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v    is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁  +  m₂u₂  = v(m₁  + m₂)

(1000 x 25)  +  (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

                                                         = 27.727(1000 + 1200)

                                                          = 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

8 0
3 years ago
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