Answer:
20.0 cm
Explanation:
Here is the complete question
The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Solution
Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.
Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.
Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m
Now, P' = 1/u + 1/v
1/u = P'- 1/v
1/u = 55.0 D - 1/0.02 m
1/u = 55.0 m⁻¹ - 1/0.02 m
1/u = 55.0 m⁻¹ - 50.0 m⁻¹
1/u = 5.0 m⁻¹
u = 1/5.0 m⁻¹
u = 0.2 m
u = 20 cm
So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.
Answer:
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Explanation:
The Nucleus: The Center of an Atom. The nucleus, that dense central core of the atom, contains both protons and neutrons. Electrons are outside the nucleus in energy levels.
Answer:
Required energy = 4758 J
Explanation:
Specific heat capacity of a material is the amount of energy required to raise the temperature of one kilogram (kg) of that material through one degree Celsius (°C).
Given data :
Specific heat capacity = c = 2440 J/kg.°C
Mass = m = 150 g = 0.15 kg
Initial temperature = 22°C
Final temperature = 35°C
Change in Temperature = ΔT = 13°C
Energy = E = ?
Using the following formula and substituting the values, we get:
E = m × c × ΔT
E = 0.15 × 2440 × 13
E = 4758 J
