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Stella [2.4K]
3 years ago
8

A 2.03 cm high insect is 1.39 m from a 131 mm focal-length lens. Where is the image? How high is it?

Physics
1 answer:
ryzh [129]3 years ago
4 0

Answer:

image is 14.47 cm behind the lens

height is 2.11 mm

Explanation:

Given data

h  = 2.03 cm

p = 1.39 m = 139 cm

focal-length f = 131 mm = 13.1 cm

to find out

Where is the image and How high is it

solution

we know focal length formula that is

1/f = 1/p + 1/q

put here value to find q

1/ 13.1 = 1/ 139  + 1 solbe/ q

q =  14.463066 cm

so image is 14.47 cm behind the lens

and

height is calculate

height / h  = - q / p

put here all value

height = -14.47 / 139 × 2.03

height = −0.211324 cm

here -ve sign show image is inverted

so height is 2.11 mm

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A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w
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Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

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Therefore

R=\rho\frac{l}{A}

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2

R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

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