When you are configuring data deduplication, you must choose a usage type for the volume you are configuring. Which of the follo
1 answer:
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Answer:
a) 2,945 mC
b) P(t) = -720*e^(-4t) uW
c) -180 uJ
Explanation:
Given:
i (t) = 6*e^(-2*t)
v (t) = 10*di / dt
Find:
( a) Find the charge delivered to the device between t=0 and t=2 s.
( b) Calculate the power absorbed.
( c) Determine the energy absorbed in 3 s.
Solution:
- The amount of charge Q delivered can be determined by:
dQ = i(t) . dt
- Integrate and evaluate the on the interval:
- The power can be calculated by using v(t) and i(t) as follows:
v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt
v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV
P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)
P(t) = -720*e^(-4t) uW
- The amount of energy W absorbed can be evaluated using P(t) as follows:
- Integrate and evaluate the on the interval:
Answer:
Part A
The mass plow rate, is approximately 97.0 lbm/s
Part B
The power used to overcome friction, is approximately 1.9 hp
Explanation:
The efficiency of the pump, η = 80%
The power input to the pump, P = 20 hp
The speed of the water through the pipe, v = 10 ft./s
The diameter of the pipe, d = 5.2 inches = 13/30 ft.
The free surface of the pool above the lake, h = 80 ft.
The density of the water, ρ = 62.4 lbm/ft.³
Part A
The mass plow rate, = Q × ρ
Where;
ρ = 62.4 lbm/ft³
Q = A × v
A = The cross-sectional area of the pipe
∴ Q = π·d²/4 × v = π × ((13/30 ft.)²)/4 × 10 ft.s ≈ 1.4748 ft.³/s
∴ The mass plow rate, ≈ 1.4748 ft.³/s × 62.4 lbm/ft.³ = 97.02752 lbm/s
The mass plow rate, ≈ 97.0 lbm/s
Part B
The power to pump the water at the given rate, =
·g·h
∴ = 97.02752 lbm/s × 32.1740 ft./s² × 80 ft. ≈ 14.1130725 Hp
≈ 14.1130725 Hp
The power output of the pump, = 0.8 × 20 hp = 16 hp
Therefore, the power used to overcome friction, =
-
∴ ≈ 16 hp - 14.1130725 Hp ≈ 1.8869275 hp
The power used to overcome friction, ≈ 1.9 hp