Answer:
(a) 20 MHz
(b) 1.025 KW
(c) 3.33 ns
(d) 33 pF
Explanation:
(a) 20,000,000 Hz = 20 x 10^6 Hz = 20 Mega Hz = <u>20 MHz</u>
(b) 1025 W = 1.025 x 10^3 W = 1.025 Kilo W = <u>1.025 KW</u>
(c) 0.333 x 10^(-8) s = 3.33 x 10^(-9) s = 3.33 nano s = <u>3.33 ns</u>
(d) 33 x10^(-12)F = 33 pico F = <u>33 pF</u>
Answer:
a) 8kW
b) $128
Explanation:
Given the coefficient of performance of the heat pump cycle to be 2.5
Energy delivered by the heat pump = 20kW
a) net power required to operate the heat pump = Energy delivered / coefficient of performance
Net power required = 20/2.5
= 8kW
b) Given the cost of electricity is $0.08 for 1kWhour
Since net power required to operate heat pump = 8kW
If the heat pump operate for 200hours, total power required for a month = 8kW×200hours = 1600kWhour
since 1kWh of electricity costs $0.08, cost of electricity used in a month when the pump operates for 200hour will be 1600kWh×$0.08 which is equivalent to $128
Answer:
A&C
Explanation:
breathing deeply is relaxing
talking with a friend can helping
Answer:
Timing Diagrams 15 pts. A 10 MHz clock that generates a 0 to 5V pulse train with a 30% duty cycle is connected to input X of a two input OR gate that has a 20nS propagation delay. The clock also goes to an inverter with a 10 ns propagation delay. The output of the inverter goes to the Y input of the OR gate. a) Draw the circuit. 2 pts. b) Plot the output of the clock for two cycles. Show times and voltages. 5 pts. c) On the same page as part (b) plot the output of the inverter. Show times and voltages. 3 pts. d) On the same page as parts (b & c) plot the output of the OR gate. Show times and voltages. 5 pts.