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Umnica [9.8K]
3 years ago
9

lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this l

ens
Physics
1 answer:
ser-zykov [4K]3 years ago
4 0
<h2>Question: </h2>

The nearpoint of an eye is 151 cm. A corrective lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this lens?

Answer:

29.96cm

Explanation:

Using the corrective lens, the image should be formed at the front of the eye and be upright and virtual.

Now using the lens equation as follows;

\frac{1}{f} = \frac{1}{v} + \frac{1}{u}   -------------(i)

Where;

f = focal length of the lens

v = image distance as seen by the lens

u = object distance from the lens

From the question;

v = -151cm        [-ve since the image formed is virtual]

u = 25cm

Rewrite equation (i) to have;

f = \frac{uv}{u+v}

Substitute the values of v and u into the equation;

f = \frac{25*(-151)}{25-151}

f = \frac{-3775}{-126}

f = 29.96cm

The focal length should be 29.96cm

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Rasek [7]

Answer:

W=173.48J

Explanation:

information we know:

Total force: F=45N

Weight: w=100N

distance: 4m

vertical component of the force: F_{y}=12N

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In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: F_{x}=Fcos\theta

vertical component: F_{y}=Fsen\theta

but from the given information we know that F_{y}=12N

so, equation these two F_{y}=Fsen\theta and F_{y}=12N

Fsen\theta =12N

and we know the force F=45N, thus:

45sen\theta=12

now we clear for \theta

sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

F_{x}=Fcos\theta

F_{x}=45cos(15.466)\\F_{x}=43.37N

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J

the work done is W=173.48J

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Displacement = 6m W

Explanation:

Given the following:

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Movement 2 = 12m E

Diatance is a scalar quantity with only magnitude and no direction. That is, in Calculating the distance moved by the locomotive, the direction of travel or movement of the object is not considered. It only measures the total amount of movement made during the Time of motion.

Therefore, total distance traveled equals :

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