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Flauer [41]
3 years ago
11

A light beam is directed parallel to the axis of a hollow cylindrical tube. When the tube contains only air, the light takes 8.7

2 ns to travel the length of the tube, but when the tube is filled with a transparent jelly, the light takes 1.82 ns longer to travel its length. What is the refractive index of this jelly?
Physics
1 answer:
monitta3 years ago
3 0

Answer:

Explanation:

velocity of light in a medium of refractive index V = V₀ / μ

V₀  is velocity of light in air and μ is refractive index of light.

time to travel in tube with air =  length of tube / velocity of light

8.72 ns = L / V₀  L is length of tube .

time to travel in tube with jelly =  length of tube / velocity of light

8.72+ 1.82  = L / V  L is length of tube .

10.54 ns = L / V

dividing the equations

10.54 / 8.72 = V₀  / V

10.54 / 8.72 =   μ

1.21 =  μ

refractive index of jelly = 1.21 .

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A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of
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Answer:

Explanation:

Given

mass of spring m=100\ gm

extension in spring x=5\ cm

downward velocity v=70\ cm/s

Position in undamped free vibration is given by

u(t)=A\cos \omega _0t+B\sin \omega _0t

where \omega _0^2=\frac{k}{m}

also \frac{k}{m}=\frac{g}{L}

\omega _0^2=\frac{k}{m}=\frac{9.8}{0.05}

\omega _0=14

u(t)=A\cos(14t)+B\sin(14t)

it is given

u(0)=0

u'(0)=70\ cm/s

substituting values we get

A=0

u(t)=B\sin (14t)

u'(t)=14B\cos (14t)

70=14B

B=\frac{10}{2}

B=5

u(t)=5\sin (14t)

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Bad White [126]

Answer:

(c) more than 500

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A projectile is launched into the air with the initial speed of vi = 40 m/s at a launch angle of 20 degrees above the horizontal
Sphinxa [80]

The range of the projectile is 188 m

Explanation:

The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction  

The path of a projectile is the combination of these two motions: see figure in attachment.

In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.

We have:

t = 5.0 s (time of fligth of the projectile)

and the horizontal velocity is constant, and it is given by

v_x = v_i cos \theta

where

v_i = 40 m/s is the initial velocity

\theta=20^{\circ} is the angle of projection

Substituting,

v_x = (40)(cos 20^{\circ})=37.6 m/s

And therefore, the range of the projectile is:

d=v_x t = (37.6)(5.0)=188 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0
3 years ago
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