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aniked [119]
3 years ago
14

What is the equation for finding the acceleration of an object moving in a straight line?

Physics
1 answer:
kiruha [24]3 years ago
4 0

Acceleration can be found using one of the following suvat equations:

v=u+at\\s=ut+\frac{1}{2}at^2\\s=vt-\frac{1}{2}at^2\\v^2-u^2=2as

Explanation:

The acceleration of an object is defined as the rate of change of velocity of the object:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to t

There are several equations used to find the acceleration of an object moving in a straight line (they are usually called suvat equation). Depending on which quantities are given in the problem, you can use one of the following equations:

v=u+at\\s=ut+\frac{1}{2}at^2\\s=vt-\frac{1}{2}at^2\\v^2-u^2=2as

where

a is the acceleration

s is the distance covered

t is the time

u is the initial velocity

v is the final velocity

Learn more about acceleration and suvat equations:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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A) 4.7 cm

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sin \theta=\frac{n \lambda}{d}

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\lambda is the wavelength

a is the distance between the slits

In this problem,

n = 5

\lambda=525 nm =5.25\cdot 10^{-7} m

a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

So we find

\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

And given the distance of the screen from the slits,

D=75.0 cm = 0.75 m

The distance of the 5th  bright fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

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Answer:

n = 1.4266

Explanation:

Given that:

refractive index of crystalline slab n = 1.665

let refractive index of fluid is n.

angle of incidence θ₁ = 37.0°

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sin \theta _ c =\frac{n}{n_{slab}}

According to Snell's law of refraction:

n sin \theta _1 = n_{slab}  \ sin \  (90- \theta_c)

At point P ; 90 - \theta _2  \leq \theta _c

\theta _2 = 90 - \theta _c

Therefore:

n \ sin \theta_1 = n_{slab} \sqrt{(1-sin^2 \theta _c)}  \\ \\ n \ sin \theta_1 = n_{slab}   \sqrt{(1- \frac{n}{n_{slab}} )}

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n = \frac{n_{slab}}{\sqrt{1+ sin^2 \theta _1 } }

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