For every 3 packages that Marcella wraps she uses 1/2 of a roll of wrapping paper if she uses 2 1/2 rolls of wrapping paper how
2 answers:
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Answer:
16 minutes
Explanation:
This is an example of a class of problems in which two quantities start with different initial values and change at different rates. In such problems, the rates of change are generally ones that cause the values to converge.
The question usually asks when the values will be the same. The generic answer is, "when the difference in rates makes up the difference in initial values."
Here the tanks differ in initial fill height by 12 -8 = 4 ft. The rates of change differ by 0.5 -0.25 = 0.25 ft/min. The more filled tank is draining faster (important), so the fill heights will converge after ...
(4 ft)/(0.25 ft/min) = 16 min
The level in the two tanks will be the same after 16 minutes.
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<em>Additional comment</em>
The oil levels at that time will be 4 ft.
You can write two equations for height:
y = 12 -0.5x . . . . . . . height in feet after x minutes (tank A)
y = 8 -0.25x . . . . . . height in feet after x minutes (tank B)
These will be equal when ...
y = y
12 -0.5x = 8 -0.25x
4 = 0.25x . . . . . . . . . . add 0.5x -8
16 = x . . . . . . . . . . . . multiply by 4 . . . . time to equal height
The graph shows when the tanks will have equal heights and when they will be drained.
Answer:
0.5 kW
Explanation:
The given parameters are;
Volume of tank = 1 m³
Pressure of air entering tank = 1 bar
Temperature of air = 27°C = 300.15 K
Temperature after heating = 477 °C = 750.15 K
V₂ = 1 m³
P₁V₁/T₁ = P₂V₂/T₂
P₁ = P₂
V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³
For ideal gas, = 5/2×R = 5/2*0.287 = 0.7175 kJ
PV = NKT
N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)
N = 9.66×10²⁴
Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles
The average mass of one mole of air = 28.8 g
Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg
∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ
The power input required = The rate of heat transfer = 149.211/(60*5)
The power input required = 0.49737 kW ≈ 0.5 kW.