Answer:
16.2 cents
Explanation:
Given that a homeowner consumes 260 kWh of energy in July when the family is on vacation most of the time.
Where Base monthly charge of $10.00. First 100 kWh per month at 16 cents/kWh. Next 200 kWh per month at 10 cents/kWh. Over 300 kWh per month at 6 cents/kWh.
For the first 100 kWh:
16 cent × 100 = 1600 cents = 16 dollars
Since 1 dollar = 100 cents
For the remaining energy:
260 - 100 = 160 kwh
10 cents × 160 = 1600 cents = 16 dollars
The total cost = 10 + 16 + 16 = 42 dollars
Note that the base monthly of 10 dollars is added.
The cost of 260 kWh of energy consumption in July is 42 dollars
To determine the average cost per kWh for the month of July, divide the total cost by the total energy consumed.
That is, 42 / 260 = 0.1615 dollars
Convert it to cents by multiplying the result by 100.
0.1615 × 100 = 16.15 cents
Approximately 16.2 cents
Answer:
add resistance to a circuit
Explanation:
It depends on the design in which it is incorporated. A fixed resistor has many uses, including, but not limited to ...
- dropping voltage
- limiting current
- contributing to a time delay
- adjusting frequency response
- eliminating (or creating) signal reflections
- acting as a fuse
- calibrating or trimming a response
- providing protection against electrical shock or ESD
- acting as a reference when measuring variable resistors
Answer:
d= 4.079m ≈ 4.1m
Explanation:
calculate the shaft diameter from the torque, \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}
Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).
r = Radius of the shaft.
T = Twisting Moment or Torque.
J = Polar moment of inertia.
C = Modulus of rigidity for the shaft material.
l = Length of the shaft.
θ = Angle of twist in radians on a length.
Maximum Torque, ζ= τ × \frac{ π}{16} × d³
τ= 60 MPa
ζ= 800 N·m
800 = 60 × \frac{ π}{16} × d³
800= 11.78 × d³
d³= 800 ÷ 11.78
d³= 67.9
d= \sqrt[3]{} 67.9
d= 4.079m ≈ 4.1m
Answer:
a) , b)
Explanation:
a) The Coefficient of Performance of the Carnot Heat Pump is:
After some algebraic handling, the temperature of the cold reservoir is determined:
b) The heating load provided by the heat pump is: