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3 years ago

The amount of diffraction that a sound wave undergoes depends on

1 answer:

5 0

The amount of diffraction of sound waves depends on the medium the sound wave travels to and the frequency. Diffraction happens as soon as it has been out of the source.

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You are designing an optical fiber scope for directing light into a confined area. You want to keep light within the fiber. Base

shepuryov [24]

Answer:

Explanation:

For entry of light into tube of unknown refractive index

sin ( 90 - 25 ) / sinr = μ , μ is the refractive index of the tube , r is angle of refraction in the medium of tube

r = 90 - C where C is critical angle between μ and body medium in which tube will be inserted.

sin ( 90 - 25 ) / sin( 90 - C) = μ

sin65 / cos C = μ

sinC = 1.33 / μ , where 1.33 is the refractive index of body liquid.

From these equations

sin65 / cos C = 1.33 / sinC

TanC = 1.33 / sin65

TanC = 1.33 / .9063

TanC = 1.4675

C= 56°

sinC = 1.33 / μ

μ = 1.33 / sinC

= 1.33 / sin56

= 1.33 / .829

μ = 1.6 Ans

3 0

3 years ago

In an open system such as a campfire matter can

eimsori [14]

In an open system such as a campfire, matter can lose particles, gain particles or exchange particles.

4 0

3 years ago

Which quantity measures the maximum distance an object on a spring moves from the equilibrium position during oscillation?

nasty-shy [4]

The appropriate response is amplitude. A measure of its change over a solitary period. There are different meanings of plentifulness, which are all elements of the extent of the distinction between the variable's outrageous qualities. In more seasoned writings, the stage is now and again called the adequacy.

5 0

3 years ago

Read 2 more answers

When using a micron gauge, if the pressure gauge falls frim 1,300 to 1,000 instantly, the vehicle needs to?

julsineya [31]

If the micron gauge pressure falls instantly, it means that the moisture of the air-conditioning is frozen, therefore you need to move the vehicle to a warmer place in order to unfreeze the moisture.

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3 years ago

A student (m = 68 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a

Gnesinka [82]

Answer:

5.7141 m

Explanation:

Here the potential and kinetic energy will balance each other

$mgh=\frac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}$

This is the initial velocity of the system and the final velocity is 0

t = Time taken = 0.04 seconds

F = Force = 18000 N

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}$

From Newton's second law

$F=ma\\\Rightarrow F=m\frac{v-u}{t}\\\Rightarrow 18000=68\frac{0-\sqrt{2gh}}{0.04}\\\Rightarrow \frac{18000}{68}\times 0.04=-\sqrt{2\times 9.81\times h}\\\Rightarrow 10.58823=-\sqrt{2\times 9.81\times h}$

Squarring both sides

$112.11061=2\times 9.81\times h\\\Rightarrow h=\frac{112.11061}{2\times 9.81}\\\Rightarrow h=5.7141\ m$

The height from which the student fell is 5.7141 m

5 0

3 years ago