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LekaFEV [45]
3 years ago
15

A police siren of frequency fsiren is attached to a vibrating platform. The platform and siren oscillate up and down in simple h

armonic motion with amplitude Ap and frequency fp. Use v for the speed of sound. Part A Find the maximum sound frequency that you would hear at a position directly above the siren. At what point in the motion of the platform is the minimum frequency heard
Physics
1 answer:
babunello [35]3 years ago
8 0

Answer:

he maximum frequency occurs when the denominator is minimum

 f’= f₀  \frac{343}{343 + v_s}

Explanation:

This is a doppler effect exercise, where the sound source is moving

           f = fo \frac{v}{v-v)s}      when the source moves towards the observer

           f ’=f_o  \frac{v}{v+v_{sy}}  Alexandrian source of the observer

the maximum frequency occurs when the denominator is minimum, for both it is the point of maximum approach of the two objects

          f’= f₀  \frac{343}{343 + v_s}

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A light ray in air enters and passes through a block of glass. What can be stated with regard to its speed after it emerges from
Alik [6]

Answer:

Speed is same as that before it entered glass.

Explanation:

Given:

A light ray enters and passes through the glass as shown in the diagram.

We have to analyze its speed.

Speed of light in air is 3\times 10^8\ ms^-^1 and speed of light in glass is 2.25\times 10^8\ ms^-^1

Whenever a light ray enters a glass block or slab there is bending of light at the interface of the two media.

So speed of light will decrease in glass medium and again it passes to the air.

So

Speed of light in air will again increase or will be equivalent to the earlier speed when it was entering the glass block.

Finally

Speed is same as that before it entered glass as it in the same medium (air).

6 0
3 years ago
Light is incident on a piece of glass in air at an angle of 33 degrees from the normal. If the index of refraction of the glass
lilavasa [31]
About 21 to 22 degrees 

as below

4 0
3 years ago
A light year is approximately 9.5 million km long. 'Barnard's Star' is 6 light years away from Earth. Calculate how many million
hammer [34]

Answer:

6 light years = 57 million km

Explanation:

Given;

A light year = 9.5 million km

To calculate how far is 6 light years;

6 light years = 6 × 1 light year = 6 × 9.5 million km

6 light years = 57 million km

7 0
3 years ago
Read 2 more answers
The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of which fun
gizmo_the_mogwai [7]

The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.

<h3>What is vertical motion of a projecile?</h3>

The vertical motion of a projectile is affected by gravity and the velocity of vertical motion given by the following formula;

Vy = Vsinθ

<h3>What is horizontal motion of a projecile?</h3>

The horizontal velocity of a projectile is given by the following formula;

Vx = Vcosθ

<h3>Direction of the motion</h3>

The direction of the motion is calculated as follows;

tanθ  = Vy/Vx

Thus, the formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.

Learn more about vertical motion here: brainly.com/question/24216590

#SPJ4

4 0
2 years ago
An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
4 years ago
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