Answer: C.
Explanation:
i choose B and got it wrong it told me the answer after on edge.
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:
a. The electric field lines are linear and perpendicular to the plates inside a parallel-plate capacitor, and always from positive plate to the negative plate. If a positive charge is released near the positive plate, then<em> it will follow a linear path towards the negative plate under the influence of electrostatic force, F = Eq</em>, where q is the charge of the particle. The electric field inside a parallel plate capacitor is constant and equal to
This can be calculated by Gauss' Law.
A positive charge always follow the electric field lines when released. Another approach is that the positive plate repels the positive charge and negative plate attracts the positive charge. Therefore, the positive charge follows a path towards the negative charge.
b. The particle moves from the higher potential to the lower potential. <em>The direction of motion is the same as the direction of the force that moves the particle, so the work done on the particle by that force is positive.</em>
Answer:
-13.18°C
Explanation:
To develop the problem it is necessary to consider the concepts related to the thermal conduction rate.
Its definition is given by the function

Where,
Q = The amount of heat transferred
t = time
k = Thermal conductivity constant
A = Cross-sectional area
The difference in temperature between one side of the material and the other
d= thickness of the material
The problem says that there is a loss of heat twice that of the initial state, that is

Replacing,




Solvinf for
,

Therefore the temprature at the outside windows furface when the heat lost per second doubles is -13.18°C
Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity. If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.
(64.2 m/s) - [ (9.8 m/s² ) x (1.5 sec) ]
= (64.2 m/s) - [ 14.7 m/s ]
= 49.5 m/s . (upward)