Answer:The awnser is 5
Explanation:Just divide all of it
Answer:
E=52000Hp.h
E=38724920Wh
E=1.028x10^11 ftlb
Explanation:
To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.
Finally, when you have the result Hp h / year you convert it to Ftlb and Wh
E=52000Hp.h
E=38724920Wh
E=1.028x10^11 ftlb
Answer:
a) the power consumption of the LEDs is 0.25 watt
b) the LEDs drew 0.0555 Amp current
Explanation:
Given the data in the question;
Three AAA Batteries;
<---- 1000mAh [ + -] 1.5 v ------1000mAh [ + -] 1.5 v --------1000mAh [ + -] 1.5 v------
so V_total = 3 × 1.5 = 4.5V
a) the power consumption of the LEDs
I_battery = 1000 mAh / 18hrs { for 18 hrs}
I_battery = 1/18 Amp { delivery by battery}
so consumption by led = I × V_total
we substitute
⇒ 1/18 × 4.5
P = 0.25 watt
Therefore the power consumption of the LEDs is 0.25 watt
b) How much current do the LEDs draw
I_Draw = I_battery = 1/18 Amp = 0.0555 Amp
Therefore the LEDs drew 0.0555 Amp current
Answer:
symbolic machine code.
Explanation:
The instructions in the language are closely linked to the machine's architecture.
Answer:
The work of the cycle.
Explanation:
The area enclosed by the cycle of the Pressure-Volume diagram of a Carnot engine represents the net work performed by the cycle.
The expansions yield work, and this is represented by the area under the curve all the way to the p=0 line. But the compressions consume work (or add negative work) and this is substracted fro the total work. Therefore the areas under the compressions are eliminated and you are left with only the enclosed area.
