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saul85 [17]
3 years ago
13

A block of mass m = 4.8 kg slides from left to right across a frictionless surface with a speed Vi=7.3m/s. It collides with a bl

ock of mass M =11.5 kg that is at rest. After the collision, the 4.8-kg block reverses direction, and its new speed is Vf=2.5m/s. What is V, the speed of the 11.5-kg block? 5.6 m/s
6.5 m/s
3.7 m/s
4.7 m/s
4.1 m/s
Physics
1 answer:
nirvana33 [79]3 years ago
7 0

Answer:

The speed of the 11.5kg block after the collision is V≅4.1 m/s

Explanation:

ma= 4.8 kg

va1= 7.3 m/s

va2= - 2.5 m/s

mb= 11.5 kg

vb1= 0 m/s

vb2= ?

vb2= ( ma*va1 - ma*va2) / mb

vb2= 4.09 m/s ≅ 4.1 m/s

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A sample of carbon dioxide (C°p,m = 37.11 J K−1 mol−1) of mass 2.80 g at 27°C is allowed to expand reversibly and adiabatically
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Answer:

182 to 3 s.f

Explanation:

Workdone for an adiabatic process is given as

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

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V₂ = 2.10 dm³ = 0.002 m³

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W =

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3 years ago
An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f
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B. 2 m/s

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4 m/s = F/m

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2.

Given that

mass m_1 = 30 kg

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\mu = 0.1

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Answer:

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