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:
Electric field, 
Explanation:
It is given that,
Magnitude of charge, 
Force experienced, 
We need to find the electric field at the origin. It is given by :
So, the electric field at the origin is 
. Hence, this is the required solution.
To calculate the specific heat capacity of an object or substance, we can use the formula
c = E / m△T
Where
c as the specific heat capacity,
E as the energy applied (assume no heat loss to surroundings),
m as mass and
△T as the energy change.
Now just substitute the numbers given into the equation.
c = 2000 / 2 x 5
c = 2000/ 10
c = 200
Therefore we can conclude that the specific heat capacity of the block is 200 Jkg^-1°C^-1
Explanation:
Initial time, t₁ = 2:30 pm
Final time, t₂ = 2:30:45
We need to find the motion of students in terms of time. Final time is 45 seconds more than the initial time.
Change in time,
Hence, this is the required solution.
Answer:
53.63 μA
Explanation:
radius of solenoid, r = 6 cm
Area of solenoid = 3.14 x 6 x 6 = 113.04 cm^2 = 0.0113 m^2
n = 17 turns / cm = 1700 /m
di / dt = 5 A/s
The magnetic field due to the solenoid is given by
B = μ0 n i
dB / dt = μ0 n di / dt
The rate of change in magnetic flux linked with the solenoid =
Area of coil x dB/dt
= 3.14 x 8 x 8 x 10^-4 x μ0 n di / dt
= 3.14 x 64 x 10^-4 x 4 x 3.14 x 10^-7 x 1700 x 5 = 2.145 x 10^-4
The induced emf is given by the rate of change in magnetic flux linked with the coil.
e = 2.145 x 10^-4 V
i = e / R = 2.145 x 10^-4 / 4 = 5.36 x 10^-5 A = 53.63 μA
