Answer:
The amount saved in a year is $ 278.6.
Explanation:
current, I = 0.275 A
Voltage, V = 120 V
cost = $ 0.20 per kWh
The energy spent in 1 year is given by
E = V I t
E = 120 x 0.275 x 365 x 24 x 3600
E = 1.04 x 10^9 J
1 kWh = 3.6 x 10^6 J
So,
E = 288.9 kWh
So, the cost is
C = $ 0.2 x 288.9 = $ 57.78
So, the amount saved = $ 336.384 - $ 57.78 = $ 278.6
Answer:
u = 1.77 m/s
Explanation:
Conservation of momentum
2250(10.3) + 3160u = (2250 + 3160)(5.32)
u = 1.77 m/s
Answer:
496.7 K
Explanation:
The efficiency of a Carnot engine is given by the equation:
![\eta = 1 - \frac{T_H}{T_L}](https://tex.z-dn.net/?f=%5Ceta%20%3D%201%20-%20%5Cfrac%7BT_H%7D%7BT_L%7D)
where:
is the temperature of the hot reservoir
is the temperature of the cold reservoir
For the engine in the problem, we know that
is the efficiency
is the temperature of the cold reservoir
Solving for
, we find:
![\frac{T_C}{T_H}=1-\eta\\T_H = \frac{T_C}{1-\eta} =\frac{298.0}{1-0.400}=496.7 K](https://tex.z-dn.net/?f=%5Cfrac%7BT_C%7D%7BT_H%7D%3D1-%5Ceta%5C%5CT_H%20%3D%20%5Cfrac%7BT_C%7D%7B1-%5Ceta%7D%20%3D%5Cfrac%7B298.0%7D%7B1-0.400%7D%3D496.7%20K)
Answer:
f = 45.13
Explanation:
Check the attached file for the solution