A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.570 diopters. Since these eyeglasses do not completely correct her vision, she must hold a newspaper 42.00 cm from her eyes in order to read it. She wears the eyeglasses 2.00 cm from her eyes. How far is her near point from her eyes

Explanation:

The refractive index of the lens is

d = 1.570 dioptres

Therefore, her focal length is

f = 1 / d

f = 1 / 1.570

f = 0.6369 m

f = 63.69cm

The distance between the newspapers and her eyes is 42cm

Distance from her eye to the glasses is 2cm

We want to calculate her near point?

So, the distance between the newspaper and the glass

do = 42 - 2 = 40cm

The distance between the glass and the virtual image formed is given as

di = f•do / (do —f)

di = 63.69 × 40 / (40 — 63.69)

di = 2547.6cm / -23.69

di = —107.54cm

The relation use is obtain from the lens equation and the negative shows that virtual image formed.

The near point from her eye is

107.54 + 2 = 109.54cm

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s