Ask question

Ask question

All categories

3 years ago

9

Which statement is true for a sound wave entering an area of warmer air?

2 answers:

Georgia [21]3 years ago

5 0

Answer:

E. The wave travels faster and with an increased wavelength.

Explanation:

As we know speed of the sound depends on the temperature of medium, higher the speed of sound more will be the temperature.

As the sound enters an area of warmer temperature, that molecules of that area posses high kinetic energy hence they vibrate faster resulting an increase in speed of sound. Now more the speed, more will be the wavelength.

Hence when sound wave enters an area of warmer air it will travel with a faster speed and with an increased wavelength.

kherson [118]3 years ago

4 0

The appropriate response is letter D. The wave ventures slower and with an expanded wavelength when a sound wave entering a range of hotter air. Hotter air implies less thick, so the wave ought to back off.

I hope the answer will help you, feel free to ask more in brainly.

You might be interested in

A motorcyclist is traveling at 58.3 mph on a flat stretch of highway during a sudden rainstorm. The rain has reduced the coeffic

Novosadov [1.4K]

Explanation:

Below is an attachment containing the solution.

8 0

4 years ago

A car of mass 1200 kg traveling westward at 30. m/s is slowed to a stop in a distance of 50. m by the car’s brakes. What was the

natka813 [3]

I believe that the answer is C. 11,000

5 0

3 years ago

Read 2 more answers

What traction of the radioisotope<br>remains in the body after one day?

r-ruslan [8.4K]

The fraction of radioisotope left after 1 day is $(\frac{1}{2})^{\frac{1}{\tau}}$ , with the half-life expressed in days

Explanation:

The question is incomplete: however, we can still answer as follows.

The mass of a radioactive sample after a time t is given by the equation:

$m(t)=m_0 (\frac{1}{2})^{\frac{t}{\tau}}$

where:

$m_0$ is the mass of the radioactive sample at t = 0

$\tau$ is the half-life of the sample

This means that the mass of the sample halves after one half-life.

We can rewrite the equation as

$\frac{m(t)}{m_0}=(\frac{1}{2})^{\frac{t}{\tau}}$

And the term on the left represents the fraction of the radioisotope left after a certain time t.

Therefore, after t = 1 days, the fraction of radioisotope left in the body is

$\frac{m(1)}{m_0}=(\frac{1}{2})^{\frac{1}{\tau}}$

where the half-life $\tau$ must be expressed in days in order to match the units.

Learn more about radioactive decay:

brainly.com/question/4207569

brainly.com/question/1695370

#LearnwithBrainly

5 0

3 years ago

A toy gun uses a spring to project a 4.5-g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the gu

marishachu [46]

Answer:

1.93 m/s

Explanation:

Parameters given:

Mass = 4.5g = 0.0045kg

Spring constant = 8.0 N/m

Length of barrel = 13 cm = 0.013m

Frictional force = 0.035N

Compression = 5.8 cm = 0.058m

First, we find the P. E. stored in the spring:

P. E. = ½*k*x²

P. E. = ½ * 8 * 0.058² = 0.013J

Then, we find the work done by the frictional force while the sphere is leaving the barrel of the gun:

Work = Force * distance

The distance here is the length of the barrel.

Work = 0.035 * 0.13 = 0.0046 J

The kinetic energy of the sphere can now be found:

K. E. = P. E. - Work done

K. E. = 0.013 - 0.0046 = 0.0084J

We can now find the speed using the formula for K. E.:

K. E. = ½*m*v²

0.0084 = ½ * 0.0045 * v²

v² = 0.0084/0.00255 = 3.733

=> v = 1.93 m/s

4 0

3 years ago

Read 2 more answers

A uniform horizontal rod of mass Mand length rotates with angular velocity about a vertical axis through its center. Attached to

Lostsunrise [7]

Answer:

o

90

Explanation:

3 0

3 years ago