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3 years ago

Wave traveling at 330 m/sec has a wavelength of 4.3 meters. What is the frequency of this wave?

Physics

1 answer:

Rasek [7]3 years ago

6 0

Answer:

76.74 Hz

Explanation:

Given:

Wave velocity ( v ) = 330 m / sec

wavelength ( λ ) = 4.3 m

We have to calculate Frequency ( f ):

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > f = v / λ

Putting values here we get:

= > f = 330 / 4.3 Hz

= > f = 3300 / 43 Hz

= > f = 76.74 Hz

Hence, frequency of sound is 76.74 Hz.

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A brass rod is 185 cm long and 1.60 cm in diameter. What force must be applied to each end of the rod to prevent it from contrac

PilotLPTM [1.2K]

Answer:

42000N

Explanation:

First you calculate how much it would contract, and secondly you then calculate the force to stretch it by that amount.

1) linear thermal expansion coef brass 19e-6 /K

∆L = αL∆T = (19e-6)(1.85)(110) = 0.00387 meter or 3.87 mm

Second part involves linear elasticity.

for brass, young's modulus is 15e6 psi or 100 GPa

cross-sectional area of rod is π(0.008)² = 0.0002 m²

F = EA∆L/L

F = (100e9)(0.0002)(0.00387) / (1.85)

F = 42000 or 42 kN

4 0

3 years ago

What is the velocity of a dog who travels south 50 meters in 9 seconds

julia-pushkina [17]

Roughly 5.6 meters per second (m/s)

Velocity is determined by distance divided by time. In this case, we divide 50 meters by 9 seconds, giving us the dog’s velocity of 5.55555 (repeating) meters per second. We can round to 5.6 m/s if needed.

5 0

3 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

7 0

4 years ago

EXPERTS/ACE/GENIUS (and people that wanna actually help)

scZoUnD [109]

Hello! Force = mass * acceleration. We can go ahead and eliminate A and B, because those answers are too small. We can multiply 4.5 (mass) * 9 (acceleration) to find the amount of force needed. 4.5 * 9 is 40.5.The amount of force needed is 40.5 N. The answer is C: 40.5 N.

8 0

3 years ago

Read 2 more answers

If the length of the air column in the test tube is 14.0 cm, what is the frequency of this standing wave?

GaryK [48]

Answer:

f = 614.28 Hz

Explanation:

Given that, the length of the air column in the test tube is 14.0 cm. It can be assumed that the speed of sound in air is 344 m/s. The test tube is a kind of tube which has a closed end. The frequency in of standing wave in a closed end tube is given by :

$f=\dfrac{nv}{4l}$

$f=\dfrac{1\times 344}{4\times 0.14}$

f = 614.28 Hz

So, the frequency of the this standing wave is 614.28 Hz. Hence, this is the required solution.

4 0

4 years ago