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Yuki888 [10]
3 years ago
5

A biconvex lens is formed by using a piece of plastic(n=1.70).

Physics
1 answer:
creativ13 [48]3 years ago
3 0

Answer:

f =17.15\ cm

Explanation:

given,

refractive index of lens, n = 1.70

Radius of curvature of front surface. R₁ = 20 cm

Radius of curvature of the back surface, R₂ = 30 cm

focal length= ?

\dfrac{1}{f}=(n-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})

    R₁ = +20 cm

    R₂ = -30 cm

    n = 1.70

\dfrac{1}{f}=(1.70-1)(\dfrac{1}{20}-\dfrac{1}{-30})

\dfrac{1}{f}=0.70 \times 0.0833

f = \dfrac{1}{0.7 \times 0.0833}

f =17.15\ cm

the focal length of the lens is equal to 17.15 cm

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(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5
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Put the value into the formula

a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}

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\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}

Put the value into the formula

\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}

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\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s

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We need to calculate the fraction of its energy that it radiates every second

\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}

\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}

Hence, The fraction of its energy that it radiates every second is 3.02\times10^{-11}.

5 0
2 years ago
Question 2 (1 point)
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Answer:

I know someone anwsered but it would be 400M

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i initial velocity (u)=10m/s

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7 0
3 years ago
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