What magnification will be produced by a lens of power –4.00 D (such as might be used to correct myopia) if an object is held 43
1 answer:
Answer:
The magnification is
Explanation:
From the question we are told that
The power of the lens is
Generally
The object distance is the negative sign is because the distance is measured in the opposite direction of incident light (i.e away )
Generally the focal length is mathematically represented as
=>
=>
converting to cm
=>
Generally from lens equation we have that
=>
=>
Generally the magnification is mathematically represented as
=>
=>
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Answer:
(a) 86.65 J
(b) 149.65 J
Solution:
As per the question:
Diameter of the pool, d = 12 m
⇒ Radius of the pool, r = 6 m
Height of the pool, H = 3 m
Depth of the pool, D = 2.5 m
Density of water,
Acceleration due to gravity, g =
Now,
(a) Work done in pumping all the water:
Average height of the pool, h =
h =
Volume of water in the pool, V =
Mass of water,
Work done is given by the potential energy of the water as:
(b) Work done to pump all the water through an outlet of 2 m:
Now,
Height, h = 2.75 + 2 = 4.75
Work done,
The gravitational force between two objects is given by:
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects
The distance of the telescope from the Earth's center is
, the gravitational force is 
and the mass of the Earth is 
, therefore we can rearrange the previous equation to find m2, the mass of the telescope:
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects
The distance of the telescope from the Earth's center is