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3 years ago

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An object 4 cm high is placed 20 cm in front of a convex lens of focal length 12 cm. What is the position and height of the imag

e? (can you please show your work)

1 answer:

aleksandrvk [35]3 years ago

6 0

It would be 12cm

hope this helps

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When the ratio of 2 variables is constant what can their relationship be described as

salantis [7]

It can be described as a constant variation

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3 years ago

What is this ???????????

babymother [125]

A, something to remember is that if the numbers are even the answer too will be even. Every time hope this helps :>

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3 years ago

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When the displacement in SHM is equal to 1/3 of the amplitude xm, what fraction of the total energy is (a) kinetic energy and (b

Nesterboy [21]

Answer:

Explanation:

Given

Displacement is $\frac{1}{3}$ of Amplitude

i.e. $x=\frac{A}{3}$ , where A is maximum amplitude

Potential Energy is given by

$U=\frac{1}{2}kx^2$

$U=\frac{1}{2}k(\frac{A}{3})^2$

$U=\frac{1}{18}kA^2$

Total Energy of SHM is given by

$T.E.=\frac{1}{2}kA^2$

Total Energy=kinetic Energy+Potential Energy

$K.E.=\frac{1}{2}kA^2 -\frac{1}{18}kA^2$

$K.E.=\frac{8}{18}kA^2$

Potential Energy is $\frac{1}{8}$ th of Total Energy

Kinetic Energy is $\frac{8}{9}$ of Total Energy

(c)Kinetic Energy is $0.5\times \frac{1}{2}kA^2$

$P.E.=\frac{1}{4}kA^2$

$\frac{1}{2}kx^2=\frac{1}{4}kA^2$

$x=\frac{A}{\sqrt{2}}$

7 0

3 years ago

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in a solar system far, far away the sun's intensity is 200 w/m2 for an inner planet located a distance r away. what is the sun's

GarryVolchara [31]

The sun's intensity for an outer planet located at a distance 6r from the sun is 5.55 W/m². The result is obtained by using the inverse square law formula.

<h3>What is the Inverse Square Law formula?</h3>

The Inverse Square Law formula describes the intensity of light is inversely proportional to the square of the distance. It can be expressed as

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

Where

I₁ = Intensity at distance 1 (W/m²)
I₂ = Intensity at distance 2 (W/m²)
d₁ = distance 1 from a light source (m)
d₂ = distance 2 from a light source (m)

Given the case the sun's intensity is 200 W/m² for an inner planet at the distance r. If an outer planet is at a distance 6r, what is the sun's intensity?

By using the inverse square law formula, the sun's intensity for an outer planet is

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

$\frac{200 }{I_{2} } = \frac{(6r)^{2} }{r^{2}}$

$\frac{200 }{I_{2} } = \frac{36r^{2} }{r^{2}}$

$I_{2} = \frac{200} {36}$

I₂ = 5.55 W/m²

Hence, the sun's intensity for a planet at a distance 6r from the sun is 5.55 W/m².

Learn more about intensity of light here:

brainly.com/question/13155277

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1 year ago

Science, who ever answers this will get a brainlest

Ganezh [65]

answer:

a.fossils of the same plants and animals are found on widely spread continents

explanation:

four fossil examples include; the mesosaurus, cynognathus, lystrosaurus, and glossopteris.

good luck :)

i hope this helps

have a nice day !!

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3 years ago

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