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sweet [91]
3 years ago
8

If a farsighted person has a near point that is 0.600 m 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 0.350 m from the person's eyes?
Physics
1 answer:
ra1l [238]3 years ago
7 0

Answer:

f = 0.84 m

Explanation:

given,

near point of the distance from eye(d_i) = 0.60 m

distance he can read (d_o)= 0.350 m

focal length of the contact lens = ?

now,

Using lens formula

\dfrac{1}{f} = \dfrac{1}{d_0} + \dfrac{1}{d_i}

\dfrac{1}{f} = \dfrac{1}{0.35} + \dfrac{1}{-0.6}

\dfrac{1}{f} = 1.19

f = \dfrac{1}{1.19}

f = 0.84 m

Hence, The focal length of the contact lens that person needed is equal to

f = 0.84 m

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The energy of the electron in Hydrogen atom can be shown to be given by En = -13.6/n^2 eV, where n represents the principal quan
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Answer:

This represents radiation in ultra-violet region .

Explanation:

Energy of the orbit where n = 3 is given as follows

E_3 =- \frac{13.6 }{3^2}

E_3 =- \frac{13.6 }{9} = -1.511 eV

Energy of the orbit where n = 1 is given as follows

E_1 =- \frac{13.6 }{1^2}

E_1 =- \frac{13.6 }{1} = 13.6 eV

Difference of [tex]E_3 and [tex]E_1 = - 1.511+ 13.6

= 12.089 eV.

The wavelength of light having this energy in nm is given by the expression as follows

Wavelength in nm = 1244 / energy in eV

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8 0
3 years ago
A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio
Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

\sum F_{x}=0

we can see there are only two forces in x, which are the weight on x and the static friction, so:

-W_{x}+f_{s}=0

when solving for the static friction we get:

f_{s}=W_{x}

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

W_{x}=mg sin \alpha

we can substitute this on our sum of forces equation:

f_{s}=mg sin \alpha

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

f_{s}=N\mu_{s}

so we can substitute this on our equation:

N\mu_{s}=mg sin \alpha

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

\sum F_{y}=0

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

N-W_{y}=0

when solving for N we get:

N=W_{y}

When seeing the free-body diagram we can determine that:

W_{y}=mg cos \alpha

so we can substitute that in the sum of y-forces equation, so we get:

N=mg cos \alpha

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

mg cos \alpha \mu_{s}=mg sin \alpha

we can divide both sides of the equation into mg so we get:

cos \alpha \mu_{s}=sin \alpha

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for \mu_{s} we get:

\mu_{s}=\frac{sin \alpha}{cos \alpha}

when simplifying this we get that:

\mu_{s}=tan \alpha

now we can solve for the angle so we get:

\alpha= tan^{-1}(\mu_{s})

and we can substitute the given value so we get:

\alpha= tan^{-1}(0.350)

which yields:

α=19.29°

which rounds to:

α=19°

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Each lunar cycle has one full moon, in which the relative positions of Earth, the sun, and the moon form in a straight line
aniked [119]

Complete Question:

Each lunar cycle has one full moon, in which the relative positions of Earth, the sun, and the moon form in a straight line. Which list represents the position of Earth, the sun, and the moon during a full moon?

Group of answer choices.

A. Earth, sun, moon

B. sun, moon, Earth

C. moon, sun, Earth

D. sun, Earth, moon

Answer:

D. sun, Earth, moon

Explanation:

A lunar eclipse is a phenomenon that occurs when the Earth comes between the Moon and the Sun thereby causing it to cover the Moon with its shadow.

Simply stated, lunar eclipse takes place when the Moon passes or moves through the Earth's shadow thereby blocking any ray of sunlight from reaching the Moon. Thus, the full moon appears deep red (blood moon).

Also, a lunar eclipse would occur only when the Sun, Earth, and Moon are closely aligned to form a straight line known as the syzygy.

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1. Total lunar eclipse.

2. Partial lunar eclipse.

3. Penumbra lunar eclipse.

Each lunar cycle has one full moon, in which the relative positions of Earth, the sun, and the moon form in a straight line. Thus, the list which represents the position of Earth, the sun, and the moon during a full moon is sun, Earth, and moon

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sergiy2304 [10]

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m = 10.0 kg is the mass of the table

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Learn more about Newton's second law:

brainly.com/question/3820012

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