A parallel circuit with two branches and an 18 volt battery. Resistor #1 on the first branch has a value of 220 ohms and resisto
1 answer:
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Answer:
Change in kinetic energy: 13479.77 J
Percentage change in kinetic energy = 99.85%
Explanation:
Change in momentum when the projectile becomes embedded in the block will be 0.
This means:
here subscript P means projectile, and subscript B means block.
Solving this equation we get:
(0.075 * 600) + (50 * 0) = (0.075 + 50) * V
Using this velocity we can compute the change in kinetic energy:
Initial kinetic energy = 0.5 * m * v^2
Initial kinetic energy = 0.5 * 0.075 * 600^2
Initial kinetic energy = 13500 J
Final Kinetic energy = 0.5 * m * v^2
Mass (m) = mass of block + mass of projectile = 50.075 kg
Final velocity v = 0.899 m/s
Final Kinetic Energy = 0.5 * 50.075 * 0.899^2
Final Kinetic Energy = 20.235 J
Change in kinetic energy = 13500 - 20.235 = 13479.77 J
Percentage change of kinetic energy = (13479.77 / 13500) * 100
Percentage change of kinetic energy = 99.85 %
Answer:
i)ω=3600 rad/s
ii)V=7059.44 m/s
iii)F=1245.8 N
Explanation:
i)
We know that angular speed given as
We know that for one revolution
θ=2π
Given that time t= 2 hr
So
ω=θ/t
ω=2π/2 = π rad/hr
ω=3600 rad/s
ii)
Average speed V
Where M is the mass of earth.
R is the distance
G is the constant.
Now by putting the values
V=7059.44 m/s
iii)
We know that centripetal fore given as
Here given that m= 200 kg
R= 8000 km
so now by putting the values
F=1245.8 N
Answer:
mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
Explanation:
The properties of water at 1 atm and thus at the saturation temperature of 100C are hfg =2256.4 kJ/kg (Table A-4). The net rate of heat transfer to the water is ˙Q=0 . 60×3 kW=1 . 8 kWNoting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, therate of evaporation of water is determined to be mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
