The Image distance and Magnification of The Image will be 30 cm and 3.
<h3>What is focal length?</h3>
The focal length of the lens, which is often expressed in millimeters, is the distance between the lens and the image sensor when the subject is in focus.
Given data;
Focal length,f=?
Image distance,v=?
Object distance,u= 10 cm
Magnification,m= 2.85
The focal length is half of the radius;
f=R/2
f=30 Cm/2
f= 15 Cm
The mirror equation is found as;

The magnification of the lens is found as;

Hence, the image distance and magnification of The image will be 30 cm and 3.
To learn more about the focal length refer;
brainly.com/question/16188698
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Answer:
kg
Explanation:
easy question to be honest
Answer:
V₁ = 6 V
, V₂ = V₃ = 3 V
Explanation:
To solve this circuit we must remember that there are two fundamental types of construction in series and parallel.
* a serial circuit there is only one path for current
in this circuit the constant current in the entire circuit and the voltage is the sum of the voltage of each term
* Parallel circuit in this there are two or more paths for the current
in this circuit the voltage is constant and the east is divided between each branch
with these principles let's analyze the proposed circuit
The DC battery is in parallel with resistor R1 and the equivalent of the other branch,
as in a parallel circuit the voltage is constant
V₁ = 6 V
in the other branch (23) it forms a series construction, where the current is constant
6 = iR₂ + iR₃
as they indicate that each resistance has the same value
6 = 2 iR
V = V₂ = V₃ = 3 V
The initial is where you are starting and the final postion is where the object ends up
Voltage = (current) x (resistance)
= (19 A) x (14 ohms) = 266 volts .
Note: Be careful using that thing !
It's dissipating
I² R = (19 A)² x (14 ohms) = 5,054 watts ! ! !
That's an awful lot of power for a blow-dryer !
The dryer is certainly not using very much of that power to run the fan.
Most of it is being used to heat air. 5 kilowatts is more power than most
toasters or microwave ovens use, so please be careful with how much of
your hair or skin you expose to that hot-air blast. You could probably cook
a meatloaf with it.