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3 years ago

A musical note has a frequency of 512 Hz. If the wavelength of the note is 0.685 m, what is the speed of the sound of that note?

Physics

1 answer:

riadik2000 [5.3K]3 years ago

5 0

Answer:

350.72 m/s

Explanation:

Formula for velocity of wave is;

v = fλ

Where;

v is speed

f is frequency

λ is wavelength

We are given;

f = 512 Hz

λ = 0.685 m

Thus;

v = 512 × 0.685

v = 350.72 m/s

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Tina is looking out a window and throws marble straight downward toward the sidewalk below at a speed of 5.00 m/s and the window

Katena32 [7]

Answer:

a) 28.1 m/s

b) 27.2 m

Explanation:

Consider the motion of the marble in downward direction taking down direction as positive

v₀ = initial velocity = 5 m/s

v = final velocity as it hits the ground = ?

a = acceleration = 9.8 m/s²

d = vertical height traveled = 39 m

Using the equation

v² = v₀² + 2 a d

v² = 5² + 2 (9.8) (39)

v = 28.1 m/s

t = time of travel = 1.90 s

y = vertical distance traveled

Using the equation

y = v₀ t + (0.5) a t²

y = (5)(1.90) + (0.5) (9.8) (1.90)²

y = 27.2 m

3 0

3 years ago

The most common example of a(n) ____ switched network is the conventional telephone system.

fenix001 [56]

CIRCUIT Switched network has the conventional telephone system as the most common example.

Circuit switching is a method of implementing a telecommunication network in which two network nodes establish a dedicated communications channel or circuit through the network before the nodes may communicate. The circuit guarantees the full bandwidth of the channel and remains connected for the duration of the communication session by functioning as though the nodes were physically connected like an electrical circuit.

In a conventional telephone system, when a call is made from one telephone to another, switches within the the telephone exchanges create a continuous wire circuit between the two units for as long as the call lasts.

4 0

4 years ago

Web Spiders and Oscillations All spiders have special organs that make them exquisitely sensitive to vibrations. Web spiders det

chubhunter [2.5K]

Complete Question

The complete quetion is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

The mass of the fly is $m_f = 11 mg = 11*10^{-3} g = 1.1*10^{-5} \ kg$

The extension of the web is $e= 4.00 \ mm = 0.004 \ m$

The spring constant is mathematically evaluated as

$k = \frac{mg}{e}$

substituting values

$k = \frac{1.1 *10^{-5} *9.8}{0.004}$

$k = 0.027 \ N/m$

The frequency of vibration is

$f = \frac{1}{2 \pi} \sqrt{\frac{k}{m} }$

substituting values

$f = \frac{1}{2 * 3.142 } \sqrt{\frac{0.027}{1.1*10^{-5}} }$

$f = 7.9 Hz$

8 0

3 years ago

You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond

Lera25 [3.4K]

Answer: Length axis f= 114.3 Hz, Width axis f=228.67 Hz

Explanation:

We are given that,

Length of tub= 1.5 m

Width of tub= 0.75 m

Sound speed= 343 m/s

Now, we are also given shower is closed.

So, frequency is given as:

f= $m* \frac{v}{2L}$

For length axis

Put v= 343 m/s, m=1 and L=1.5 m

f= 1 * $\frac{343}{2*1.5}$

f= 114.3 Hz

For next resonant frequency, m=2

f= 2* $\frac{343}{2*1.5}$

f= 228.67

For width axis

Put v= 343 m/s, m=1 and L= 0.75 m

f= 1* $\frac{343}{2*0.75}$

f= 228.67 Hz

For next frequency, m=2

f= 2* $\frac{343}{2*0.75}$

f= 457.34 Hz

6 0

4 years ago

The cue ball is deflected so that it makes an angle of 30.0° with its original direction of travel.

Alexus [3.1K]

Answer:

(a) θ= 43.89°

(b) $v_{1} = 1.88\sqrt{3} \frac{m}{s}$

$v_{2} = 6.79 \frac{m}{s}$

Explanation:

Ball 1:

$u_{1} = 7.52\frac{m}{s}$

Ball 2:

$u_{2} = 0\frac{m}{s}$

As the collision is elastic, it means that kinetic energy and momentum are conserved. Following that, we apply the law of conservation of energy and momentum:

$\frac{1}{2} m_{1}u_{1}^{2} = \frac{1}{2} m_{1}v_{1}^{2} + \frac{1}{2} m_{2}v_{2}^{2}$

and

$m_{1}u_{1} = m_{1}v_{1} + m_{2}v_{2}$

Where u is the velocity before the collision, and v are the velocities after the collision. Both previous equations can be simplified as:

$u_{1}^{2} = v_{1}^{2} + v_{2}^{2}$

and

$u_{1} = v_{1} + v_{2}$

This because the two balls have the same mass. We know that the cue ball is deflected and makes an angle of 30°. From conservation in x-direction, we get::

$56.55 = v_{1} ^{2} + v_{2} ^{2}$

and

$u_{1} = v_{1}cos(30) + v_{2}cos(\theta)$

Solving we get:

$(\frac{2u_{1}-\sqrt{3}v_{1} }{2cos(\theta)})^{2} = 56.55-v_{1} ^{2}$

From conservation in y-direction, we get:

$0 = v_{1}sin(30) - v_{2}sin(\theta)$

From this, we solve this equation system and get the answers. Remember to add 30° to the angle obtained.

8 0

3 years ago