The ratio of the kinetic energy of the block/bullet system immediately after the collision to the initial kinetic energy of the bullet is 0.78 %.
<h3>Final velocity of the block/bullet system</h3>
Apply the principle of conservation of energy to determine the final velocity of the block/bullet system.
K.E = P.E
¹/₂mv² = mgh
¹/₂v² = gh
v² = 2gh
v = √2gh
where;
- h is the maximum height reached by the system
- v is the initial velocity of the system
v = √(2 x 9.8 x 1.1)
v = 4.64 m/s
<h3>Initial velocity of the bullet</h3>
Apply the principle of conservation of linear momentum.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
- u₁ is the initial velocity of the bullet
- u₂ is the initial velocity of the block
- v is the final velocity after collision
- m₁ is mass bullet
- m₂ is mass of block
(0.0075)u₁ + (0.95)(0) = 4.64(0.0075 + 0.95)
0.0075u₁ = 4.4428
u₁ = 4.4428/0.0075
u₁ = 592.37 m/s
<h3>Initial kinetic energy of the bullet</h3>
K.Ei = ¹/₂m₁u₁²
K.Ei = ¹/₂(0.0075)(592.37)²
K.Ei = 1,315.88 J
<h3>Final kinetic energy of the block/bullet system</h3>
K.Ef = ¹/₂(m₁ + m₂)v²
K.Ef = ¹/₂(0.0075 + 0.95)(4.64)²
K.Ef = 10.31 J
<h3>Ratio of final kinetic energy to initial kinetic energy</h3>
= K.Ef/K.Ei x 100%
= (10.31 / 1,315.88) x 100%
= 0.78 %
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1.7 Btu
1 watt = 3.41214 Btu/h
1watt * 1h = 3.41214 Btu/h * h
1 = 3.41214 Btu/ (watt*h)/
0.5 watt * h = 0.5 watt*h * 3.41214 Btu/(watt*h) = 1.706 Btu
Answer:
231.98 mL
Explanation:
V₁ = Initial volume = 205 mL
V₂ = Final volume
P₁ = Initial pressure = 712 mmHg
P₂ = Final pressure = 750 mmhg (STP)
T₁ = Initial temperature = -44 °C = 229.15 K
T₂ = Final temperature = 273.15 K (STP)
From Combined gas law
Volume of the argon at STP is 231.98 mL
Answer:
A= 203 KJ
B= 54 Kg
Explanation:
The initial specific volumes and internal energies are obtained from A-12 for a given pressure and state. The enthalpy of the refrigerant in the supply line is determined using the saturated liquid approximation for the given temperature with data from A-11. The mass that has entered the tank is:
Δm = m₂ – m₁
= V(1/α₂ – 1/α₁)
= 0.05 (1/0.0008935 – 1/ 0.025645)Kg
= 54Kg
The heat transfer is obtained from the energy balance:
ΔU= 
+ 
m₂u₂ – m₁u₂ = +
= m₂u₂ – m₁u₁ –
= V/α₂u₂ - V/α₁u₁ –
=(0.05/0.0008935 . 116.72 – 0.05/0.025645 . 246.82 – 54.108.28) Kj
= 203 KJ
