Ask question

Ask question

All categories

2 years ago

11

1) A sound wave with a frequency of 300 hertz is traveling through a medium at a speed of 320 meters/second. What is its wavelen

gth?
2) A sound wave has a wavelength of 0.450 meters. If its speed in cold air is 330 meters/second, what is the wave's frequency?
a.0.890hertz
b.100hertz
c.230hertz
d.733

2 answers:

timurjin [86]2 years ago

5 0

1. The wavelength is the ratio of the wave's speed to its frequency in hertz or 1/s. This is shown below,
λ = s / f = (320 m/s) / (300 1/s) = 1.07 m
The wavelength is approximately 1.07 m.

2. The frequency is the ratio between speed and the wavelength,
f = (330 m/s) / 0.45 m = 733.33 hertz

Kazeer [188]2 years ago

3 0

1. The wavelength is the ratio of the wave's speed to its frequency in hertz or 1/s. This is shown below,

λ = s / f = (320 m/s) / (300 1/s) = 1.07 m

The wavelength is approximately 1.07 m.

2. The frequency is the ratio between speed and the wavelength,

f = (330 m/s) / 0.45 m = 733.33 hertz

You might be interested in

A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is in

Vikki [24]

49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

$\boxed{u \ or \ v_i = initial \ velocity}$
$\boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}$
$\boxed{a = acceleration \ (constant)}$
$\boxed{d = distance \ travelled}$

The formula we follow for this problem are as follows:

$\boxed{ \ v^2 = u^2 + 2ad \ }$

a = acceleration (in m/s²)
u = initial velocity
v = final velocity
d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

$\boxed{ \ 0 = v^2 + 2ad \ }$

$\boxed{ \ v^2 = -2ad \ }$

Both sides are divided by -2d, we get $\boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }$

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

$\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }$

Here d' is the stopping distance that we want to look for.

$\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }$

We crossed out 2 in above and below.

$\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }$

We multiply both sides by d.

$\boxed{ \ v^2 d' = 49.0v^2 d \ }$

We crossed out v^2 on both sides.

$\boxed{\boxed{ \ d' = 49.0d \ }}$

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

<h3>Learn more</h3>

Determine the acceleration of the stuffed bear brainly.com/question/6268248
Particle's speed and direction of motion brainly.com/question/2814900
About the projectile motion brainly.com/question/2746519

Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

6 0

2 years ago

Read 2 more answers

The constant pressure molar heat capacity, C_{p,m}C p,m , of nitrogen gas, N_2N 2 , is 29.125\text{ J K}^{-1}\text{ mol}^{-1

antoniya [11.8K]

Answer:

Explanation:

Constant pressure molar heat capacity Cp = 29.125 J /K.mol

If Cv be constant volume molar heat capacity

Cp - Cv = R

Cv = Cp - R

= 29.125 - 8.314 J

= 20.811 J

change in internal energy = n x Cv x Δ T

n is number of moles , Cv is molar heat capacity at constant volume , Δ T is change in temperature

Putting the values

= 20 x 20.811 x 15

= 6243.3 J.

3 0

2 years ago

Jade and her roommate Jari commute to work each morning, traveling west on I-10. One morning Jade left for work at 6:45 A.M., bu

Veronika [31]

Answer:

Jari

Explanation:

The question requires to know who is traveling faster. This is done by comparing the gradients. The steeper the slope (high gradient), the faster the speed and vice versa.

From Jari's line, the starting point is (0, 0) and another point is (6, 7)

The gradient being change in y to change in x

Change in y=7-0=7

Change in x=6-0=6

Slope is 7/6

For Jade, first point is (0, 10) then another point is (6, 16)

Change in y=16-10=6

Change in x=6-0=6

Slope is 6/6=1

Clearly, 7/6 is greater than 6/6 or 1 hence Jari is faster than Jade

3 0

2 years ago

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the crit

Dennis_Churaev [7]

Complete Question:

A beam of white light is incident on the surface of a diamond at an angle $\theta_{a}$ , since the index of refraction depends on the light's wavelength, the different colors that comprise white light will spread out as they pass through the diamond. For example, the indices of refraction in diamond are $n_{red} = 2.410$ for red light and $n_{blue} = 2.450$ for blue light. Thus, blue light and red light are refracted at different angles inside the diamond. The surrounding air has $n_{air} = 1.000$ .

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the critical angle θcrit, the light will not be refracted out into the air, but instead it will be totally internally reflected back into the diamond. Find θcrit. Express your answer in degrees to four significant figures.

Answer:

$\theta_{crit} = 24.09^{0}$

Explanation:

Only the blue refracted ray is related to the critical angle in this question

$n_{air} = 1.000$

$n_{blue} = 2.450$

The relationship between the critical angle( $\theta_{crit}$ ), $n_{air}$ and $n_{blue}$ can be given as $sin \theta_{crit} = \frac{n_{air} }{n_{blue} }$

$sin \theta_{crit} = \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} 0.4082^{0}\\ \theta_{crit} = 24.09^{0}$

6 0

2 years ago

If one of the satellites is at a distance of 20,000 km from you, what percent accuracy in the distance is required if we desire

Lesechka [4]

<span>It is quite straightforward to convert an uncertainty to a percent uncertainty. We can divide the amount of uncertainty by the original amount and then multiply by 100%.

(2 m / 20,000,000 m) X 100% = 0.00001%

The percent uncertainty is 0.00001%.

The percent accuracy is the 100% - percent uncertainty.
The percent accuracy = 100% - 0.00001% = 99.99999%

The percent accuracy is 99.99999%.</span>

8 0

2 years ago