C. since the the heat from the heater is going to the child in <u>waves</u>, it’s<u> radiating </u>
Answer:
27.95[kW*min]
Explanation:
We must remember that the power can be determined by the product of the current by the voltage.
where:
P = power [W]
V = voltage [volt]
I = amperage [Amp]
Now replacing:
![P=110*8.47\\P=931.7[W]](https://tex.z-dn.net/?f=P%3D110%2A8.47%5C%5CP%3D931.7%5BW%5D)
Now the energy consumed can be obtained mediate the multiplication of the power by the amount of time in operation, we must obtain an amount in Kw per hour [kW-min]
![Energy = 931.7[kW]*30[days]*10[\frac{min}{1day} ]=279510[W*min]or 27.95[kW*min]](https://tex.z-dn.net/?f=Energy%20%3D%20931.7%5BkW%5D%2A30%5Bdays%5D%2A10%5B%5Cfrac%7Bmin%7D%7B1day%7D%20%5D%3D279510%5BW%2Amin%5Dor%2027.95%5BkW%2Amin%5D)
Question four bulbs A,B,C and D are connected in a circuit shown in the figure below, the letters X, Y and Z represent three switches. Which switch is used to operate switch A separately?
Answer: x
An object distance is
presented as s = 5f and we know that the mirror equation relates the image
distance to the object distance and the focal length.
The mirror equation is
1/f = 1/s + 1/s’ where the variable f stands for
the focal length of the mirror. Variable (s)
represents the distance between the mirror surface and the object and the
variable <span>(s’) represents the distance between the mirror surface and
the image. </span>
In addition, a concave mirror
will have a positive focal length (f) and a convex mirror will have a negative
focal length (f).
Now, we then have 1/f = 1/5f
+ 1/s’ which is s’ = 5f/4
Then we get the magnification
ratio that expresses the size or amount of magnification or reduction of the
object or image and to get the magnification, we use this equation: M= s’/s
M= 5f/4x5f
s’ = 1/4s
Therefore, the image height
is one fourth of the object height
