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Vaselesa [24]
3 years ago
7

A 0.180-kilogram car traveling at 0.80 m/s to the right collides with a 0.100-kilogram cart intially at rest. The carts lock tog

ether upon collison. Calculate the final velocity of the carts
Physics
1 answer:
prohojiy [21]3 years ago
6 0

0.51 m/s to the right.

<h3>Explanation</h3>

The two carts lock together. As a result, the collision is inelastic. Kinetic energy will not conserve. Still, momentum conserves.

What's the momentum <em>p</em> of the two carts?

Before the collisions:

  • p_1 = m_1 \cdot v_1 = 0.180 \times 0.80 = 0.144 \; \text{kg} \cdot \text{m}\cdot \text{s}^{-1};
  • p_2 = m_2 \cdot v_2 = 0.
  • Sum of momentum: p = p_1 + p_2 = 0.144\; \text{kg} \cdot \text{m}\cdot \text{s}^{-1}.

Momentum conserves. As a result, p(\text{after collision}) = p(\text{before collision}) = 0.144 \; \text{kg} \cdot \text{m}\cdot \text{s}^{-1}

Velocity is the same for the two carts after the collision. Let v denote that velocity.

p(\text{after collision}) = m_1\cdot v+ m_2 \cdot v = (m_1 + m_2) \cdot v.

v = \dfrac{p(\text{after collision})}{m_1 + m_2} \\ \phantom{v} = \dfrac{0.144}{0.180 + 0.100} \\ \phantom{v} = 0.51 \; \text{m}\cdot \text{s}^{-1}.

Direction of the movement will stay the same. Both cars are now moving to the right.

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Carolina (600 N) runs up the stadium (10 meters high) in 9.14 s. How much power did Carolina expend?
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2 years ago
A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
3 years ago
2. A 50 kg diver is standing on the edge of a 15 m high cliff. What is his potential energy?
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Answer:

3675 J

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Gravitational Potential Energy = \frac{1}{2} × mass × g × height

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Mass = 50 kg

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3 0
3 years ago
A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.50 x 105 P
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From the question we are told that:

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Answer:

10

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