Answer:
B. 180 million joules
Explanation:
Apply the formula for heat transfer given as;
Q=m*c*Δt where
Q = electrical energy consumed by the heater in joules
m= mass of air in the chamber in kg
c= specific heat of air in joules per kg degrees Celsius
Δt= change in temperatures in degrees Celsius
Given in the question;
m= 1200 kg
c= 1000 J/°C /kg
Δt = 180°-30°= 150° C
Substitute values in the equation to get Q as;
Q=m*c*Δt
Q= 1200 * 1000* 150
Q= 180000000 joules
Q = 180 million joules
<u>The correct answer option is B : 180 million joules.</u>
Yes that’s correct . It’s safer cause of the current are even
As there are 10 V, for Vp1, that is the peak-voltage of the source:
Then, transformer's theory says that the relation of transformations is:
V1/V2=a
Where V1 is the voltage in the primary and V2 in the secondary.
V1=14.14 V
V2=8.55 V
a=1.65
Then, with the 8.5 V, we find the real peak-voltage, taking in account that in the diodes we have a drop of 0.7 V each, so:
8.5 -1.4=7.1 V
And this will be called VpL
Now we proceed to calculate the mean voltage:
Where Vr is the ripple voltage, we asume that is 1 V
So, Vmean = 6.6 V
Then we have
Vmean/R= I mean
We have that R=1000 Ohm
Imedia=6.6 V/1000 Ohm
Imedia=6.6 mAmps
Finally, we can calculate the capacitor:
C=Q/Vr
C=Imean/(Vr*2f)
Where f is 60Hz
C=6.6mA/(1V*120)
C=5.5 uFarads
Therefore:
C=5.5 uFarads that works at 12 V
Answer:
Small businesses contribute to local economies by bringing growth and innovation to the community in which the business is established. Small businesses also help stimulate economic growth by providing employment opportunities to people who may not be employable by larger corporations.
Answer:
Explanation:
Lets do this in python, our function will import an array of double numbers, then it sort the array out, drop the first and last items, aka highest and lowest score. Finally, it averages the last 3 numbers by calculate the sum and divide by the number of items
def calculate_score(scores):
scores.sort() # sort it so that the lowest is at index 0 and the highest is at last index
drop_highest_lowest = scores[1:-1] # this will drop the lowest and highest number
average_score = sum(drop_highest_lowest)/len(drop_highest_lowest)
return average_score
