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ra1l [238]
3 years ago
12

A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a

frictionless surface and when the carts collide they stick together. What is the ratio of the initial kinetic energy of the two-cart system to the final kinetic energy of the two-cart system?
Physics
1 answer:
IgorLugansk [536]3 years ago
5 0

Answer:1.5

Explanation:

Given

mass of first  cart m_1=6 kg

initial Velocity u_1=3 m/s

mass of second cart m_2=3 kg

u_2=0 m/s

In the absence of External Force we can conserve momentum

m_1u_1+m_2u_2=(m_1+m_2)v

v=\frac{m_1u_1+m_2u_2}{m_1+m_2}

v=\frac{6\times 3+3\times 0}{6+3}

v=2 m/s

Final kinetic Energy of two masses

K.E._2=\frac{1}{2}(m_1+m_2)v^2

K.E._2=\frac{1}{2}\cdot (3+6)\cdot (2)^2

K.E._2=18 J

Initial Kinetic Energy

K.E._1=\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2

K.E._1=\frac{1}{2}6\times 3^2+0

K.E._1=27 J

ratio =\frac{K.E._1}{K.E._2}=\frac{27}{18}=1.5

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3 years ago
5) [Honors]A seagull, ascending straight upward at 5.2 m/s, drops a shell when it is 12.5m above the ground. (A)
jolli1 [7]

Answer:

(B) 13.9 m

(C) 1.06 s

Explanation:

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v₀ = 5.2 m/s

y₀ = 12.5 m

(A) The acceleration in free fall is -9.8 m/s².

(B) At maximum height, v = 0 m/s.

v² = v₀² + 2aΔy

(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)

y = 13.9 m

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3 0
3 years ago
6. During an impact time casting 5 x 10-45 a gulf club exerts an average impar
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Answer:

2.5 × 10-⁴¹ Ns

Explanation:

Impulse

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I = 25 × 10-⁴² Ns

I = 2.5 × 10-⁴¹ Ns

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2 years ago
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