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Lilit [14]
3 years ago
12

If a nearsighted person has a far point df that is 3.50m from the eye, what is the focal length f1 of the contact lenses that th

e person would need to see an object at infinity clearly?
If a farsighted person has a near point that is 0.600m from the eye, what is the focal length f2 of the contact lenses that the person would need to be able to read a book held at 0.350m from the person's eyes?

Physics
1 answer:
olga55 [171]3 years ago
3 0

Answer:

f1= -350cm or -3.5m

f2= 22.1cm or 0.221m

Explanation:

A person is nearsighted when the person's far point is less than infinity. A diverging lens is normally used to correct this eye defect. A diverging lens has a negative focal length as seen in the solution attached.

Farsightedness is when a person's near point is farther than 25cm. This eye defect is corrected using a converging lens. The focal length of a converging lens is positive. This is evident in the solution attached. The near point is also referred to as the least distance of distinct vision.

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(a) The electron will move towards the wire.

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\vec{F} = q\vec{v} \times \vec{B}

According to the above formula, the direction of the force the wire applies to the electron can be found by right-hand rule.

Since the electron has a negative charge, the direction of the force is towards the wire.

(b) The proton will veer to the right.

The direction of the magnetic field is the same as the previous part. The proton has a positive charge, and coming from above. The direction of its velocity is downwards. The magnetic field above the wire is pointed into the page. Using the right-hand rule, the magnetic force on the proton is directed to the right, with respect to us.

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3 years ago
You push the ball with aforce of 22.8N which induces a -2.3N frictional force. What is the net force while you push?
inna [77]
The net force is 22.8-2.3 or 20.5 N
5 0
3 years ago
A ball is thrown horizontally from the top of a 55 m building and lands 150 m from the base of the building. Ignore air resistan
PtichkaEL [24]

Answer:

a) t =3.349 s

b) V_x,i = 44.8 m/s

c) V_y,f = 32.85 m/s

d)  V = 55.55 m/s

Explanation:

Given:

- Total throw in x direction x(f) = 150 m

- Total distance traveled down y(f) = 55 m

Find:

a) How long is the rock in the air in seconds.  

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

                                 y(f) = y(0) + V_y,i*t + 0.5*g*t^2

- V_y,i = 0 (horizontal throw)

                                 55 = 0 + 0 + 0.5*(9.81)*t^2

                                 t = sqrt ( 55 * 2 / 9.81 )

                                 t =3.349 s

- Use the second equation of motion in x direction:

                                 x(f) = x(0) + V_x,i*t

                                 150 = 0 + V_x,i*3.349

                                  V_x,i = 150 / 3.349 = 44.8 m/s

- Use the first equation of motion in y direction:

                                 V_y,f = V_y,i + g*t

                                 V_y,f = 0 + 9.81*3.349

                                 V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

                                 V^2 = V_y,f^2 + V_x,i^2

                                 V = sqrt (32.85^2 + 44.8^2)

                                 V = 55.55 m/s

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