Answer:
it is not possible to place the wires in the condui
Explanation:
given data
total area = 2.04 square inches
wires total area = 0.93 square inches
maximum fill conduit = 40%
to find out
Can it is possible place wire in conduit conduit
solution
we know maximum fill is 40%
so here first we get total area of conduit that will be
total area of conduit = 40% × 2.04
total area of conduit = 0.816 square inches
but this area is less than required area of wire that is 0.93 square inches
so we can say it is not possible to place the wires in the conduit
Answer:
(iv) second law of thermodynamics
Explanation:
The Clausius inequality expresses the second law of thermodynamics it applies to the real engine cycle.It is defined as the cycle integral of change in entropy of a reversible system is zero. It is nothing but mathematical form of second law of thermodynamics . It also states that for irreversible process the cyclic integral of change in entropy is less than zero
Answer:
450,000m = 450km = 4.5E5
32,600,000W = 32.6MW = 3.26E7
59,700,000,000cal = 59.7Gcal = 5.97E10
0.000000083s = 83ns = 8.3E-8
35,000Ω = 35kΩ = 3.5E4
Explanation:
Giga = 1,000,000,000
Mega = 1,000,000
kilo = 1,000
unit = 1
deci = .1
centi = .01
milli = .001
micro = .000001
nano = .0000000001
pico = .000000000001
You should be able to look at these and convert between them in seconds if you want to pursue anything in engineering.
Answer:
The reaction at support B
Rb= 235440N
The reaction at support C
RC= 29430N
Explanation : check attachment
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