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Mama L [17]
2 years ago
6

Calculate the ratio of change in the mass of the molecules of a gas to the initial mass, if its

Engineering
1 answer:
kirill115 [55]2 years ago
3 0

<h2>Answer:</h2>

<h3>Required Answer is as follows :-</h3>

  • \sf \dfrac{P _{i}}{ P_{f}} = \dfrac{3}{4}

  • 3/4 = (⅓ × m₁/Volume × V²)/(⅓ × m₂/Volume) × (V/2)²

  • ¾ = m₁/Volume × (V)²/m₂/Volume × (V/2)²

  • ¾ = m₁ × (V)²/m₂ × (V²/4)

  • ¾ = m₁/m₂ × 4

  • m₂ = 4 × 4/3 = 16/3m₁

\bold{   } 

<h3>Now,</h3>
  • ∆m = m₂ - m₁
  • ∆m = 16m₂/3 - m₁ = 13m₁/3
  • Ratio = (13m₁/3)/ m₁
  • Ratio = 13/3

Ratio = 13:3

\bold{   } 

<h3>Know More :-</h3>

Mass => It is used to measure it's resistance to its acceleration. SI unit of mass is Kg.\bold{   } 

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If a hoist lifts a 4500lb load 30ft in 15s, the power delivered to the load is a) 18.00hp b) 9000hp c) 16.36hp d) None of the ab
12345 [234]

Answer:

Explanation:

load = 4500lb                   lift height= 30 ft

time =15 s

velocity=\frac{30}{15} ft/s

velocity=2 ft/s

power = force\times velocity

power={4500}\times2

power= 9000 lb ft/s

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2 years ago
Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,
Lyrx [107]

Answer:

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0

Explanation:

v_0 = Initial velocity of ball A

v_A=v_0\cos45^{\circ}

v_B = Initial velocity of ball B = 0

(v_A)_n' = Final velocity of ball A

v_B' = Final velocity of ball B

e = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

Coefficient of restitution is given by

e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

Adding the above two equations we get

2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0

\boldsymbol{\therefore v_B'=0.6364v_0}

(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0

From the conservation of momentum along the plane of contact we have

(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}

v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0.

5 0
3 years ago
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