Answer:
<h2> $1.50</h2>
Explanation:
Given data
power P= 2 kW
time t= 15 min to hours = 15/60= 1/4 h
cost of power consumption per kWh= 10 cent = $0.1
We are expected to compute the cost of operating the heater for 30 days
but let us computer the energy consumption for one day
Energy of heater for one day= 2* 1/4 = 0.5 kWh
the cost of operating the heater for 30 days= 0.5*0.1*30= $1.50
<u><em>Hence it will cost $1.50 for 30 days operation</em></u>
Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.


Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
Answer:
Answer is D. 8.04 x 10^4 J
Explanation:
1. D
2. A
3. D
4. B
5. C
6. B
7. D
8. C
9. B
10. D
All correct i promise you that
Check the attached file for the answer.
Answer:
A. 112 J
Explanation:
KE = ½mv² = ½(0.14)40² = 112 J