Answer: a=9.8*10^-10s
b=9.8*10^-13s
c=1.7*10^-8s
d=5.57*10^-4s
Explanation:
a) given 980 ps
Expected answer is 980 * 10^-12
Therefore, 980ps = 9.8*10^-10s
b) given 980 fs
Expected answer is 980 * 10^-15
Therefore, 980fs = 9.8*10^-13s
c) given 17 ns
Expected answer is 17 * 10^-9
Therefore, 17ns = 1.7*10^-8s
d) given 577 μs
Expected answer is 577 * 10^-6
Therefore, 577μs = 5.57*10^-4s
a=9.8*10^-10s
b=9.8*10^-13s
c=1.7*10^-8s
d=5.57*10^-4s
Answer:
Total charge flow through the cooker is 21600 C
Explanation:
As we know that the current flow through the cooker is given by Ohm's law
here it is given as




now the charge flow through it is given as

total time is t = 8 hours


Answer:
The maximum energy stored in the combination is 0.0466Joules
Explanation:
The question is incomplete. Here is the complete question.
Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Energy stored in a capacitor is expressed as E = 1/2CtV² where
Ct is the total effective capacitance
V is the supply voltage
Since the capacitors are connected in series.
1/Ct = 1/C1+1/C2+1/C3
Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF
1/Ct = 1/11.7 + 1/21.0 + 1/28.8
1/Ct = 0.0855+0.0476+0.0347
1/Ct = 0.1678
Ct = 1/0.1678
Ct = 5.96μF
Ct = 5.96×10^-6F
Since V = 125V
E = 1/2(5.96×10^-6)(125)²
E = 0.0466Joules
Answer:
V_infinty=98.772 m/s
Explanation:
complete question is:
The following problem assume an inviscid, incompressible flow. Also, standard sea level density and pressure are 1.23kg/m3(0.002377slug/ft3) and 1.01imes105N/m2(2116lb/ft2), respectively. A Pitot tube on an airplane flying at standard sea level reads 1.07imes105N/m2. What is the velocity of the airplane?
<u>solution:</u>
<u>given:</u>
<em>p_o=1.07*10^5 N/m^2</em>
<em>ρ_infinity=1.23 kg/m^2</em>
<em>p_infinity=1.01*10^5 N/m^2</em>
p_o=p_infinity+(1/2)*(ρ_infinity)*V_infinty^2
V_infinty^2=9756.097
V_infinty=98.772 m/s
Answer:
circuit sketched in first attached image.
Second attached image is for calculating the equivalent output resistance
Explanation:
For calculating the output voltage with regarding the first image.

![Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V](https://tex.z-dn.net/?f=Vout%20%3D%205%20%5Cfrac%7B2000%7D%7B5000%7D%5B%2F%5Btex%5D%3C%2Fp%3E%3Cp%3E%5Btex%5DVout%20%3D%205%20%5Cfrac%7B2000%7D%7B5000%7D%5C%5CVout%20%3D%205%20%5Cfrac%7B2%7D%7B5%7D%20%3D%202%20V)
For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.
so.

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.
if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:




so.
