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Evgesh-ka [11]
3 years ago
6

A string that is 3.6 m long is tied between two posts and plucked. The string produces a wave that has a frequency of 320 Hz and

travels with a speed of 192 m/s. How many full wavelengths of the wave fit on the string?
Physics
1 answer:
marin [14]3 years ago
5 0

To solve this problem it is necessary to apply the concepts related to wavelength depending on the frequency and speed. Mathematically, the wavelength can be expressed as

\lambda = \frac{v}{f}

Where,

v = Velocity

f = Frequency,

Our values are given as

L = 3.6m

v= 192m/s

f= 320Hz

Replacing we have that

\lambda = \frac{192}{320}

\lambda = 0.6m

The total number of 'wavelengths' that will be in the string will be subject to the total length over the size of each of these undulations, that is,

N = \frac{L}{\lambda}

N = \frac{3.6}{0.6}

N = 6

Therefore the number of wavelengths of the wave fit on the string is 6.

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3 years ago
The frequency of a microwave is 1.2 x 10^9 hertz. what is the wavelength of the given problem.
Olenka [21]

Answer:

0.25 m

Explanation:

Electromagnetic waves consist of oscillations of the electric and the magnetic field, oscillating in a plane perpendicular to the direction of motion the wave.

All electromagnetic waves travel in a vacuum always at the same speed, the speed of light, whose value is:

c=3.0\cdot 10^8 m/s

Microwave is an example of electromagnetic waves.

The relationship between wavelength and frequency for an electromagnetic wave is:

\lambda=\frac{c}{f}

where

\lambda is the wavelength

c=3.0\cdot 10^8 m/s  is the speed of light

f is the frequency

For the microwave in this problem,

f=1.2\cdot 10^9 Hz

So its wavelength is

\lambda=\frac{3.0\cdot 10^8}{1.2\cdot 10^9}=0.25 m

7 0
4 years ago
A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh
Studentka2010 [4]

Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446 rad/sec^2

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius r=\frac{d}{2}=\frac{1.63}{2}=0.815m

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

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(b) We know that linear speed is given by

v=r\omega =0.815\times 8.362=6.815m/sec

(c) We have given final angular velocity \omega _f=675rev/min

And \omega _i=79.9rev/min

Time t = 63 sec

Angular acceleration is given by \alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2

(d) Change in angle is given by

\Theta =\frac{1}{2}(\omega _i+\omega _f)t=\frac{1}{2}(675+79.9)\times 1.05=396.22rev

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Paha777 [63]

Answer:

D

Explanation:

Work = Distance x Mass

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