Answer:
Yes, it is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water that is entering at 20 °C because this temperature (20 °C) of the external cooling water is less than the saturation temperature of steam which is which is 45.81 °C, and heated by a boiler; as a result of this condition, coupled with the assumption that the turbine, pump, and interconnecting tube are adiabatic, and the condenser exchanges its heat with the external cooling river water, it possible to maintain a pressure of 10 kPa.
Answer:
Time taken for the capacitor to charge to 0.75 of its maximum capacity = 2 × (Time take for the capacitor to charge to half of its capacity)
Explanation:
The charging of a capacitor/the build up of its voltage follows an exponential progression and is given by
V(t) = V₀ [1 - e⁻ᵏᵗ]
where k = (1/time constant)
when V(t) = V₀/2
(1/2) = 1 - e⁻ᵏᵗ
e⁻ᵏᵗ = 0.5
In e⁻ᵏᵗ = In 0.5 = - 0.693
-kt = - 0.693
kt = 0.693
t = (0.693/k)
Recall that k = (1/time constant)
Time to charge to half of max voltage = T(1/2)
T(1/2) = 0.693 (Time constant)
when V(t) = 0.75
0.75 = 1 - e⁻ᵏᵗ
e⁻ᵏᵗ = 0.25
In e⁻ᵏᵗ = In 0.25 = -1.386
-kt = - 1.386
kt = 1.386
t = 1.386(time constant) = 2 × 0.693(time constant)
Recall, T(1/2) = 0.693 (Time constant)
t = 2 × T(1/2)
Hope this Helps!!!
Answer:
Explanation:
R1 = 1 k; R2 = 1k; R3 = 3k; R4 = 6k; R5 = 5k; R6 = 6k and R7 = 2k
Vt = 428V
The series and parallel circuit combination is as follow:
(R6║R7 + R5) + R4 + R3║R2 + R1
(6*2/6 + 2) + 5 = 13/ k
(13/2*6/13/6 + 6) = 78/31k
78/3 + 3 = 171/3 = 57k
57k║R2 = 57k║1k = 57/58k + R1 = (57/58 + 1)k = 115/58k = 2k
It = Vt/2k = 428/2000 = 0.2A
∴ I1 = 0.2A
I1 = I2 + I3
Using current divider rules to obtain I2 and I3
∴ I2 = I1 X (I2/I2 + I3) = 0.2 X ( 1/4) = 0.05A
and I3 = I1 X (I3/I2 + I3) = 0.2 X (3/4) = 0.15A
I3 = I4 + I5, using current divider
I4 = I3 X (I4/I4 + I5) = 0.15 X (6/6 + 5) = 0.08
I5 = 0.15 - 0.08 = 0.07A
I5 = I6 + I7, using current divider
I6 = I5 X (I6/I6 + I7) = 0.07 X (6/6 + 2) = 0.05A
I7 = 0.07 - 0.05 = 0.02A
Step 1- put shoe on
Step 2- grab the laces and cross them
Step 3- once you crossed them pick them back up and put one underneath one.
Step 4- tighten it then make two bunnies
Step 5- put on bunny over the other then loop it around
Step 6 pull the loops and bam you now know how to tie your shoes
Answer:
A) After 2.5 milliseconds
Explanation:
Given that :
In a CAN bus, there are three computers: Computer A, Computer B, and Computer C.
Length in time of the message sent are 2 milliseconds long
As the computer start sending the message; the message are being sent at a message duration rate of 2 milliseconds
250 microseconds later, Computer B starts to send a message; There is a delay of 250 microseconds = 0.25 milliseconds here at Computer B
and 250 microseconds after that, Computer C starts to send a message
Similarly; delay at Computer C = 0.25 milliseconds
Assuming 
is the retransmit time for Computer A to retransmit its message, Then :
= 
= 2 milliseconds + 0.25 milliseconds + 0.25 milliseconds
= 2.5 milliseconds
Thus; the correct option is A) After 2.5 milliseconds
