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Alecsey [184]
3 years ago
8

Using the lensmaker's formula (equation (5) of your lab manual), calculate the index of refraction of the acrylic lens. You shou

ld use the f_are you calculated in part (1) above instead of the value for the focal length of the concave lens that you measured. Remember that the focal length of a concave lens is negative so in this case, f = -f_are.
Physics
1 answer:
uysha [10]3 years ago
8 0

Answer:

  n = 1 + R / f

Explanation:

The equation of the constructor is optical is

          1 / f = 1 / p + 1 / q

where f is the focal length, p and q are the distance to the object and image, respectively

The exercise tells us that it is a concave lens with focal length fo, in these lenses the focal length is negative. The relationship to calculate the focal length is

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

where is n₀ the refractive index of the medium that surrounds the lens in this case it is air with n₀ = 1, you do not indicate the type of lens, but the most used lens is the concave plane, in this case R₂ = ∞, so which 1 / R₂ = 0, let's substitute

         1 / f = (n-1) / R₁

         n - 1 = R₁ / f

let's calculate

         n = 1- R₁ / f

remember that the radius of curvature is negative, so the equation is

         n = 1 + R / f

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A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.
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1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

1)

The motion of the shell is a projectile motion, so we  can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

y=(u sin \theta)t-\frac{1}{2}gt^2 (1)

where:

u sin \theta is the initial vertical velocity of the shell, with u=156 ft/s and \theta=49.0^{\circ}

g=32 ft/s^2 is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

x=(ucos \theta)t

where u cos \theta is the initial horizontal velocity of the shell.

We can re-write this last equation as

t=\frac{x}{u cos \theta} (1b)

And substituting into (1),

y=xtan\theta -\frac{1}{2}gt^2 (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of \alpha=-41.0^{\circ} from the horizontal, so we can write the position (x,y) of the hill as

y=x tan \alpha (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

xtan\alpha = xtan \theta - \frac{1}{2}gt^2

Substituting (1b) into this equation,

xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0

Which has 2 solutions:

x = 0 (origin)

and

tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft

So, the distance d down the hill at which the shell strikes the hill is

d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m

2)

In order to find how long the mortar shell remain in the air, we can use the equation:

t=\frac{x}{u cos \theta}

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x = 1322 ft is the final position of the shell when it strikes the hill

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Substituting these values into the equation, we find the time of flight of the shell:

t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s

3)

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s

Instead, the vertical component of the velocity is given by

v_y=usin \theta -gt

And substituting at t = 12.9 s (time at which the shell strikes the hill),

v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s

Therefore, the  final speed of the shell is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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