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4 years ago

What is pitch related to in terms of sound? Does it affect the speed? If so, how?

Physics

1 answer:

Juli2301 [7.4K]4 years ago

5 0

Pitch is the impression the listener gets of the frequency of the sound.

The speed of the sound is not related to its pitch/frequency.

If the speed and frequency were related, that would be a real problem. Bands, orchestras, and choirs could not exist ! All the instruments in the orchestra could play a note together, at the same time. But then the higher instruments ... the flute, trumpet, violins, high guitar strings and high piano keys ... would travel to you fast, and the lower instruments ... the trombone, tuba, double bass, bass drum, low guitar strings and the low piano keys ... would travel to you slow. They all played the note at the same time, but by the time you heard it, it would be all smeared out ... every instrument arriving at your ear at a different time !

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Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib

Alexus [3.1K]

Answer:

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = $\pi d^{2} /4$ = (3.142 x $(2*10^{-3} )^{2}$ )/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J

c) The peak electric field is given as

E = $\sqrt{2I/ce_{0} }$

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

$e_{0}$ = 8.85 x 10^9 m kg s^-2 A^-2

E = $\sqrt{2*318.2/3*10^8*8.85*10^9}$ = 1.55 x 10^-8 v/m

6 0

4 years ago

____________ are used to calculate the distance a continent has moved in a year. a. Space satellites c. Laser beams b. Mirrors d

Law Incorporation [45]

Space satellites, laser beams, mirrors are used to calculate the distance a continent has moved in a year.

Therefore, your correct answer would be "all of the above".

4 0

3 years ago

A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position

Nesterboy [21]

The position of the object at time t =2.0 s is 6.4 m.

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

$v_x= \frac{dx}{dt}$

Write an equation for x.

$dx=v_xdt\\ x=\int v_xdt$

Substitute the equation for vₓ =2t² in the integral.

$x=\int v_xdt\\ =\int2t^2dt\\ =\frac{2t^3}{3} +C$

Here, the constant of integration is C and it is determined by applying initial conditions.

When t =0, x = 1. 1m

$x= \frac{2t^3}{3} +C\\ x_0=1.1\\ x= (\frac{2t^3}{3} +1.1)m$

Substitute 2.0s for t.

$x= (\frac{2t^3}{3} +1.1)m\\ =\frac{2(2.0)^3}{3} +1.1\\ =6.43 m$

The position of the particle at t =2.0 s is 6.4m

5 0

3 years ago

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If

arlik [135]

Answer:

Part a)

$a = 1260.3 m/s^2$

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

$v_f^2 - v_i^2 = 2 a d$

$v_1^2 - 0^2 = 2(9.81)(4.0)$

$v_1 = 8.86 m/s$

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - v_2^2 = 2(-9.81)(2.00)$

$v_2 = 6.26 m/s$

Part a)

Average acceleration is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

$a = 1260.35 m/s^2$

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0

3 years ago

If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?

sdas [7]

Data:
$f_{2} = 42 Hz$
n (Wave node)
V (Wave belly)
L (Wave length)
The number of bells is equal to the number of the harmonic emitted by the string.

$f_{n} = \frac{nV}{2L}$

Wire 2 → 2º Harmonic → n = 2

$f_{n} = \frac{nV}{2L}$
$f_{2} = \frac{2V}{2L} 
$
$2V = f_{2} *2L$
$V = \frac{ f_{2}*2L }{2}$
$V = \frac{42*2L}{2}$
$V = \frac{84L}{2}$
$V = 42L$

Wire 1 → 1º Harmonic or Fundamental rope → n = 1

$f_{n} = \frac{nV}{2L}$
$f_{1} = \frac{1V}{2L}$
$f_{1} = \frac{V}{2L}$

If, We have:
V = 42L
Soon:
$f_{1} = \frac{V}{2L}$
$f_{1} = \frac{42L}{2L}$
$\boxed{f_{1} = 21 Hz}$

Answer:

The fundamental frequency of the string:
21 Hz

7 0

3 years ago

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