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2 years ago

An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.

Physics

1 answer:

Pachacha [2.7K]2 years ago

4 0

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

$u -v= d$

$u=71+26=97\ cm$

We need to calculate the focal length

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}$

$\dfrac{1}{f}=-\dfrac{71}{2522}$

$f=-35.52\ cm$

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value in to the formula

$\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}$

$\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}$

$\dfrac{1}{v}=-\dfrac{124}{2755}$

$v=-22.21\ cm$

We need to calculate the magnification

Using formula of magnification

$m=\dfrac{v}{u}$

$m=\dfrac{22.21}{95}$

$m=0.233$

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

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2 years ago

A wooden plaque is in the shape of an ellipse with height 30 centimeters and width 22 centimeters. Find an equation for the elli

Volgvan

Answer:

Explanation:

height of Ellipse $=30 cm$

i.e. $2 a=30$

Width of Ellipse $=22 cm$

i.e. $2 b=22$

Equation of a vertical Ellipse is

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

$a=15 ,b =11$

$\frac{x^2}{11^2}+\frac{y^2}{15^2}=1$

at $y=4 cm$

$\frac{x^2}{11^2}+\frac{4^2}{15^2}=1$

$x=\frac{11}{15}\times \sqrt{15^2-4^2}$

$x=10.6 cm$

5 0

2 years ago

Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot

m_a_m_a [10]

Answer:0.27

Explanation:

Given

One worker Pushes with force $F_1=430 N$

other Pulls it with a rope of rope $F_2=360 N$

mass of crate $m=290 kg$

both forces are horizontal and crate slides with a constant speed

Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

$f_r=F_1+F_2$