1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
The stopping distance d of a car after the brakes are applied varies directly as the square of the speed r. If a car travelling
Crank
270/70^2  = x/80^2

Cross multiply

270 (6400) = 4900x   

x = 270(6400)/4900

352 and 32/49 feet

Hope this helps


4 0
3 years ago
A projectile is launched horizontally from a 20-m tall edifice with a vox of 25 m/s. How long will it take for the projectile to
NISA [10]

Answer:

a) First let's analyze the vertical problem:

When the projectile is on the air, the only vertical force acting on it is the gravitational force, then the acceleration of the projectile is the gravitational acceleration, and we can write this as:

a(t) = -9.8m/s^2

To get the vertical velocity we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

where v0 is the initial vertical velocity because the object is thrown horizontally, we do not have any initial vertical velocity, then v0 = 0m/s

v(t) = (-9.8m/s^2)*t

To get the vertical position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + p0

where p0 is the initial position, in this case is the height of the edifice, 20m

then:

p(t) = (-4.9m/s^2)*t^2+ 20m

The projectile will hit the ground when p(t) = 0m, then we need to solve:

(-4.9m/s^2)*t^2+ 20m = 0m

20m = (4.9m/s^2)*t^2

√(20m/ (4.9m/s^2)) = t = 2.02 seconds

The correct option is a.

b) The range will be the total horizontal distance traveled by the projectile, as we do not have any horizontal force, we know that the horizontal velocity is 25 m/s constant.

Now we can use the relationship:

distance = speed*time

We know that the projectile travels for 2.02 seconds, then the total distance that it travels is:

distance = 2.02s*25m/s = 50.5m

Here the correct option is a.

c) Again, the horizontal velocity never changes, is 25m/s constantly, then here the correct option is option b. 25m/s

d) Here we need to evaluate the velocity equation in t = 2.02 seconds, this is the velocity of the projectile when it hits the ground.

v(2.02s) =  (-9.8m/s^2)*2.02s = -19.796 m/s

The velocity is negative because it goes down, and it matches with option d, so I suppose that the correct option here is option d (because the sign depends on how you think the problem)

4 0
3 years ago
A person, with a mass of 50.0 kg, stands on a weighing scale in a lift which is moving
svet-max [94.6K]

Answer:

D

Explanation:

Because I just had that answer

6 0
3 years ago
A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5
harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0
3 years ago
Uma carga puntiforme de + 3,0uC é colocada em um ponto P de um campo elétrico gerado por uma partícula eletrizada com carga desc
expeople1 [14]

Responda:

1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5

Explicação:

Dado o seguinte:

Carga (q) = 3uC = 3 × 10 ^ -6C

Força elétrica (Fe) = 18N

Intensidade do campo elétrico (E) =?

1)

Lembre-se:

Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)

Fe = qE; E = Fe / q

E = 18N / (3 × 10 ^ -6C)

E = 6N / 10 ^ -6C

E = 6 × 10 ^ 6NC ^ -1

2)

Lembre-se:

E = kQ / r ^ 2

E = intensidade do campo elétrico

Q = carga de origem

r = distância de espera = 30cm = 30/100 = 0,3m

K = 9,0 × 10 ^ 9

6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2

9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09

Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9

Q = 0,06 × 10 ^ (6-9)

Q = 0,06 × 10 ^ -3

Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC

7 0
4 years ago
Other questions:
  • Which food provides the body with energy in the form of carbohydrates
    7·2 answers
  • A rock weighing 20 n (mass = 2 kg) is swung in a horizontal circle of radius 2 m at a constant speed of 6 m/s. what is the centr
    13·2 answers
  • A boat having a length 3m and breadth 2m is floating on a
    12·1 answer
  • If you push a book against a wall hard enough it will not slide down even though gravity is pulling it. Use what you know about
    12·1 answer
  • Isaac drop ball from height og 2.0 m, and it bounces to a height of 1.5 m what is the speed before and after the ball bounce?
    6·1 answer
  • What tension must a 50.0 cm length of string support in order to whirl an attached 1,000.0 g stone in a circular path at 5.00 m/
    8·1 answer
  • 47. Calculate the final steady temperature obtained when
    7·1 answer
  • Forces between objects act when the objects are in direct contact or when they are not touching. Can you think of any forces tha
    9·2 answers
  • When elements bond together or when bonds of compounds are broken and form a new substance
    14·2 answers
  • What are the two conditions for equilibrium of a rigid body? how would this experiment demonstrate their validity?.
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!