Explanation:
(i)
O is the object and I is the image.
The image formed is enlarged and it is erect. So the magnification will be positive (+) and greater than 1.
Refer above image. 1
(ii)
O is the object and I is the image.
The image formed is diminished and erect. So the magnification will be positive (+) and less than1.
Refer above image. 2
(iii)
The image will be formed as the 2F on the other side of the lens and it will be of same of the object.
Its Kinetic, hope this helps you
efficiency = (useful energy transferred ÷ energy supplied) × 100
It's easy to use this formula, but we have to know both the useful energy and the energy supplied. The drawing doesn't tell us the useful energy, so we have to find a clever way to figure it out. I see two ways to do it:
<u>Way #1:</u>
We all know about the law of conservation of energy. So we know that the total energy coming out must be 250J, because that's how much energy is going in. The wasted energy is 75J, so the rest of the 250J must be the useful energy . . . (250J - 75J) = 175J useful energy.
(useful energy) / (energy supplied) = (175J) / (250J) = <em>70% efficiency</em>
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<u>Way #2: </u>
How much of the energy is wasted ? . . . 75J wasted
What percentage of the Input is that 75J ? . . . 75/250 = 30% wasted
30% of the input energy is wasted. That leaves the other <em>70%</em> to be useful energy.
From the concept of optics on a curvature of a spherical mirror, the proportion for which the focal length is equivalent to half the radius of curvature is fulfilled. Mathematically this is

Here,
f = Focal Length
R = Radius
Rearranging to find the radius we have,

Replacing with our values,
R = 2(13.8cm)
R = 27.6cm
Therefore the radius of the spherical surface from which the mirror was made is 27.6cm
Answer:
72km
Explanation:
30 mins --> 30 x 60 s = 1800 s
Distance --> Speed x Time
= 40m/s x 1800s
= 72 000 m
= 72 km (1km is 1000m)