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Oduvanchick [21]
2 years ago
6

Calculate the radius of curvatuire of the concave lens based on the measured focal length.

Physics
1 answer:
Roman55 [17]2 years ago
8 0

Answer:

 R₁ = (n -1) f

Explanation:

In geometric optics the focal length and the radius of curvature are related, for the case of a lens

          1 / f = (n₂-n₀) (1 / R₁ - 1 / R₂)

where f is the focal length, n₂ is the refractive index of the material, n₀ is the refractive index of the medium surrounding the material, R₁ and R₂ are the radius of curvature of each of the material's

In our case, the most common is that the lens is in the air, so n1 = 1, in many cases one of the surfaces is flat, so its radius of curvature R₂ = ∞.

     1 / f = (n-1) (1 / R₁)

we look for the radius of curvature R₁

    1 / R₁ = 1 / f (n-1)

     R₁ = (n -1) f

With this expression we can find the radius of curvature of a concave-plane lens

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A 7.7 kg sphere makes a perfectly inelastic collision with a second sphere initially at rest. The composite system moves with a
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15.4 kg.

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From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m').................... Equation 1

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.

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B. the stars to come back to the same positions in the sky.

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3 years ago
A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas
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Answer:

change in momentum, \Delta p=7.65 \,kg.m.s^{-1}

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Average Force, F=144.3396\,N

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  • F_y=72.1698\,N

Explanation:

Given:

angle of kicking from the horizon, \theta= 30^{\circ}

velocity of the ball after being kicked, v=18 m.s^{-1}

mass of the ball, m=0.425\, kg

time of application of force, t=5.3\times 10^{-2}\,s

We know, since body is starting from the rest

\Delta p=m.v.....................(1)

\Delta p=0.425\times 18

\Delta p=7.65 \,kg.m.s^{-1}

Now the components:

\Delta p_x= 7.65\times cos 30^{\circ}

\Delta p_x= 6.6251 \,kg.m.s^{-1}

similarly

\Delta p_y= 7.65\times sin 30^{\circ}

\Delta p_y= 3.825 \,kg.m.s^{-1}

also, impulse

I=F\times t.........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

F\times t=m.v

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F_x=144.3396\times cos 30^{\circ}

F_x=125.0018\,N

&

F_y=144.3396\times sin 30^{\circ}

F_y=72.1698\,N

6 0
3 years ago
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