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
Answer:
Explanation:
Given
average speed of Phelps 
total distance 
he swims first 100 m at an average speed if 
so time taken is 
suppose
is the time taken to swim remaining half
average velocity is 



so velocity in the second half is

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:
0.12 K
Explanation:
height, h = 51 m
let the mass of water is m.
Specific heat of water, c = 4190 J/kg K
According to the transformation of energy
Potential energy of water = thermal energy of water
m x g x h = m x c x ΔT
Where, ΔT is the rise in temperature
g x h = c x ΔT
9.8 x 51 = 4190 x ΔT
ΔT = 0.12 K
Thus, the rise in temperature is 0.12 K.