Answer:
Daily from the combustion chamber of a wood-burning heater.
<h3>
Explanation:</h3>
- As a wood stove heats up, it radiates heat through the walls and top of the stove
- This radiant heat warms the immediate area and can be carried into other parts of the home via the home's natural airflow.
- Electric or convection-powered fans can help circulate this heat to warm a larger area.
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Answer: C. There is no relationship
Explanation:
At the various temperatures, the flow rates do not seem to show any relation as similar temperatures can yield different flow rates.
The flow rates neither appear to generally increase nor decrease as a result of a decrease or an increase in temperature which means that there is a lack of a positive, negative and curvilinear relationship.
There is simply no relationship.
Answer:
1) the mean surface temperature of earth.
2) precipitation and sunshine
3) air and wind movement.
Answer:
I can share a link to answer since the code is to long to be contained here. the code is on github
https://github.com/arafatm/edu_coursera_machine_learning_1_foundations/blob/master/code/02.01.predicting.house.prices.ipynb
Answer:
A) 30 mH
B ) 10-ohm
Explanation:
resistor = 10-ohm
Inductor = 30mH ( l )
L = inductance
R = resistance
r = internal resistance
values of the original Inductors
Note : inductor = constant time (t) case 1
inductor + 10-ohm resistor connected in series = constant time ( t/2) case2
inductor + 10-ohm resistor + 30 mH inductance in series = constant time (t) case3
<em>From the above cases</em>
case 1 = time constant ( t ) = L / R
case 2 = Req = R + r hence time constant t / 2 = L / R + r therefore
t = 
case 3 = Leq = L + l , Req = R + r . constant time ( t )
hence Z =
= t
A) Inductance
To calculate inductance equate case 1 to case 3
=
= L / 10 = (L + 30) / ( 10 + 10 )
= 20 L = 10 L + 300 mH
10L = 300 m H
therefore L = 30 mH
B ) The internal resistance
equate case 1 to case 2
= 
R + r = 2 R therefore ( r = R ) therefore internal resistance = 10-ohm