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ipn [44]
3 years ago
10

Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?

Physics
1 answer:
Lerok [7]3 years ago
7 0

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

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adell [148]

Answer:

19.3

Explanation:

Assuming we have to find Specific gravity of gold.

As we know that specific gravity is defined as the ratio of weight of the object and weight of the water displaced by the object

so it is given by

specific gravity = weight of the object/weight of the water displaced

now we have

weight of the object = (density)(volume)g

weight of object = (19.3)(0.55)g

now weight of the liquid displaced is given by

weight of water displaced = (1 g/cm^3)(0.55ml)g

now we have

specific gravity = (19.3×0.55)/(1×0.55)

specific gravity= 19.3

8 0
3 years ago
What is the reverse of vaporization?
Naddik [55]
Answer: condensation.

Vaporization is the pass from liquid state to gaseous state.

Then the reverse is the transformation from gaseous state to liquid state.

That is called condensation.

When the water vaporizes the liquid transforms into vapor which goes to the atmosphere. When the water vapor of the atmosphere condensates liquid water is formed. You can see condensation when you have a glass with cold water and drops of water form in the exterior of the glass: those drops are liquid water that formed when the vapor of the air that surrounds the glass cools due to the lower temperature of the surface of the glass.


8 0
3 years ago
A projectile is launched at an angle above the
gtnhenbr [62]
The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.

The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.

v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s

Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.

The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.

Combining this together we get:
(1) vx=40m/s and vy=10m/s

7 0
3 years ago
GIVING BRAINLIEST TO THE CORRECT ANSWER!!
Basile [38]
The answer is A. locations by the ocean typically do not get as cold in the winter or as hot in the summer as locations that are located inland.
6 0
2 years ago
Read 2 more answers
The specific heat of a certain type of metal is 0.128 J/(g⋅∘C).0.128 J/(g⋅∘C). What is the final temperature if 305 J305 J of he
Makovka662 [10]

Answer:

45.3°C

Explanation:

Heat gained = mass × specific heat × increase in temperature

q = mC (T − T₀)

Given C = 0.128 J/g/°C, m = 94.0 g, q = 305 J, and T₀ = 20.0°C:

305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)

T = 45.3°C

6 0
3 years ago
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