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:
35°C
Explanation:
q = mCΔT
2130 J = (0.200 kg) (710 J/kg/°C) (T − 20.0°C)
T = 35°C
Answer:
A
Explanation:
voltage between A and C is equal battery's voltage.
The random variable in this experiment is a Continuous random variable.
Option D
<u>Explanation</u>:
The continuous random variable is random variable where the data can take infinite variables. For example random variable is taken for measuring "speed of automobiles" on the highways. The radar instrument depicts time taken by automobile in particular what speed. They are the generalization of discrete random variables not the real numbers as a random data is created. It gives infinite sets of all possible outcomes. It is obvious that outcomes of the instrument depend on some "physical variables" those are not predictable as depends on the situation.
<u>Given that</u>
mass (m) = 1300 Kg ,
height (h) = 1500 m
Determine the potential energy ?
P.E = m × g × h
= 1300 × 9.81 × 1500
= 19129500 Joules
= 19129.5 KJ
