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3 years ago

What is the frequency of a sound wave commonly called?

1 answer:

3 0

Frequency of a sound wave is commonly referred to as pitch. That is the specialized name for frequency of a sound wave.

Just remember it as pitch.

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Consider a horizontal, uniform board of weight 125 N and length 4 m that is supported by vertical chains at each end. A person w

Norma-Jean [14]

Answer:

b. 375 N

Explanation:

System of forces in balance

ΣFy = 0 Equation (1)

Forces acting on the board

T₁: Tension in the left chain , vertical and upward

T₂ = 250 N : Tension in the right chain , vertical and upward

W₁ = 125 N : Weight of the board , vertical and downward

W₂ = 500 N : Weight of the person , vertical and downward

Calculation of the T₁

We apply the Equation (1)

ΣFy = 0

T₁+T₂-W₁-W₂ = 0

T₁ = -T₂+W₁+W₂

T₁ = -250 N+ 125 N+ 500 N

T₁ = 375 N

5 0

3 years ago

Please help me plsss!

alexira [117]

Answer:

Explanation:

3 0

2 years ago

The wavelength of red helium-neon laser light in air is 632.8 nm.(a) What is its frequency?(b) What is its wavelength in glass t

inn [45]

(a) $4.74 \cdot 10^{14}Hz$

The frequency of a wave is given by:

$f=\frac{v}{\lambda}$

where

v is the wave's speed

$\lambda$ is the wavelength

For the red laser light in this problem, we have

$v=c=3\cdot 10^8 m/s$ (speed of light)

$\lambda=632.8 nm=632.8\cdot 10^{-9} m$

Substituting,

$f=\frac{3\cdot 10^8 m/s}{632.8 \cdot 10^{-9} m}=4.74 \cdot 10^{14}Hz$

(b) 427.6 nm

The wavelength of the wave in the glass is given by

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda_0 = 632.8\cdot 10^{-9} m$ is the original wavelength of the wave in air

n = 1.48 is the refractive index of glass

Substituting into the formula,

$\lambda=\frac{632.8\cdot 10^{-9}m}{1.48}=427.6\cdot 10^{-9}m=427.6 nm$

(c) $2.02\cdot 10^8 m/s$

The speed of the wave in the glass is given by

$v=\frac{c}{n}$

where

$c = 3\cdot 10^8 m/s$ is the original speed of the wave in air

n = 1.48 is the refractive index of glass

Substituting into the formula,

$v=\frac{3\cdot 10^8 m/s}{1.48}=2.02\cdot 10^8 m/s$

5 0

3 years ago

D section a only still need help on this

SIZIF [17.4K]

Acceleration means speeding up, slowing down, or changing direction. The graph doesn't show anything about direction, so we just have to examine it for speeding up or slowing down ... any change of speed.

The y-axis of this graph IS speed. So the height of a point on the line is speed. If the line is going up or down, then speed is changing.

Sections a, c, and d are all going up or down. Section b is the only one where speed is not changing. So we can't be sure about b, because we don't know if the track may be curving ... the graph can't tell us that. But a, c, and d are DEFINITELY showing acceleration.

5 0

3 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

3 0

3 years ago