Answer:
amount of energy = 4730.4 kWh/yr
amount of money = 520.34 per year
payback period = 0.188 year
Explanation:
given data
light fixtures = 6
lamp = 4
power = 60 W
average use = 3 h a day
price of electricity = $0.11/kWh
to find out
the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66
solution
we find energy saving by difference in time the light were
ΔE = no of fixture × number of lamp × power of each lamp × Δt
ΔE is amount of energy save and Δt is time difference
so
ΔE = 6 × 4 × 365 ( 12 - 9 )
ΔE = 4730.4 kWh/yr
and
money saving find out by energy saving and unit cost that i s
ΔM = ΔE × Munit
ΔM = 4730.4 × 0.11
ΔM = 520.34 per year
and
payback period is calculate as
payback period = 
payback period = 
payback period = 0.188 year
If the boat's velocity is 18m/sec relative to the water in the river and not the shore, it would need to be added the river speed of 2.5m/sec to get a total of 20.5m/sec. The 20.5m/sec would then be the total velocity of the boat relative to the shore. From personal experience, I know that when one runs with the tide, one is adding the tide flow speed to one's boat speed (what it would be in neutral waters) to get a sometimes much faster speed.
-3 m/s
---------
per min
oh I think 8m/s to 3m/s to 0m/s
idk probably -0.08
Answer:
3.5 cm
Explanation:
mass, m = 50 kg
diameter = 1 mm
radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m
L = 11.2 m
Y = 2 x 10^11 Pa
Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3
= 7.85 x 10^-7 m^2
Let the wire is stretch by ΔL.
The formula for Young's modulus is given by


ΔL = 0.035 m = 3.5 cm
Thus, the length of the wire stretch by 3.5 cm.
Answer:
4.41 W
Explanation:
P = IV, V = IR
P = V² / R
Given that P = 0.0625 when V = 1.50:
0.0625 = (1.50)² / R
R = 36
So the resistor is 36Ω.
When the voltage is 12.6, the power consumption is:
P = (12.6)² / 36
P = 4.41
So the power consumption is 4.41 W.