" A sound wave moving through air consists of alternate regions of high pressure and low pressure. If the frequency of the sound is increased, what happens, if anything, to the distance between successive high-pressure regions, and why

Answers

Answer 1

Answer:

Answer: wavelength will reduce

Explanation: The region of low pressure is know as the rarefraction region while the region of high pressure is the compression region.

The distance between 2 successive rarefraction or compression region is known as the wavelength.

Now the question is concerned about what an increase in frequency will cause to wavelength.

The speed of sound in air is a constant and it is approximately 343 m/s.

But recall that v = fλ

By assuming a fixed value for speed (v), we have that

k = fλ

Hence, f = k/ λ

This implies that at a fixed wave speed, the wavelength and frequency have an inverse relationship.

An increase in frequency will bring about a reduction in wavelength.

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Hearing the siren of an approaching fire truck, you pull over to side of the road and stop. As the truck approaches, you hear a tone of 460 Hz; as the truck recedes, you hear a tone of 410 Hz. How much time will it take to jet from your position to the fire 5.00 km away, assuming it maintains a constant speed?
Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.
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Two cars, a Porsche Boxster convertible and a Toyota Scion xB, are traveling at constant speeds in the same direction. Suppose, instead, that the Boxster is initially 170 m behind the Scion. The speed of the Boxster is 24.4 m/s and the speed of the Scion is 18.6 m/s. How much time does it take for the Boxster to catch the Scion

If a person’s weight is W on the surface of the earth, calculate what it would be, in terms of W, at the surface of (a) the moon;
(b) Mars;
(c) Jupiter.

Answers

Answer:b

Explanation:

Air trapped in the cooling system could create undesirable areas of combustion heat buildup in the engine called _____.

Answers

Answer:

Air pockets.

Explanation:

Air pockets in the cooling system are bubbles of air trapped within the lines (hoses and pipes) of the cooling system. This air bubbles enter the cooling system usually during the process of filling the radiator coolant fluid (usually water), or replacing the water pump or the radiator hose during repairs or servicing of the cooling system. The trapped air prevent pressure movement that is needed by the coolant to move the heat generated from the engine cylinder, resulting in heat build up. The solution is to "bleed" the engine through the radiator lid or some air release valves.

Suppose that a particle accelerator is used to move two beams of particles in opposite directions. In a particular region, electrons move to the right at 6020 m/s and protons move to the left at 1681 m/s. The particles are evenly spaced with 0.0476 m between electrons and 0.0662 m between protons. Assuming that there are no collisions and that the interactions between the particles are negligible, what is the magnitude of the average current in this region

Answers

Answer:

$2.429783984* 10^(-14)\ A$

Explanation:

Velocity of electron = 6020 m/s

Velocity of proton = 1681 m/s

Electron space = 0.0476 m

Proton space = 0.0662 m

e = Charge of particle = $1.6* 10^(-19)\ C$

Number of electrons passing per second

$n_e=(6020)/(0.0476)\n\Rightarrow n_e=126470.588$

Number of protons passing per second

$n_p=(1681)/(0.0662)\n\Rightarrow n_p=25392.749$

Current due to electrons

$I_e=n_ee\n\Rightarrow I_e=126470.588* 1.6* 10^(-19)\n\Rightarrow I_e=2.0235* 10^(-14)\ A$

Current due to protons

$I_p=n_pe\n\Rightarrow I_p=25392.749* 1.6* 10^(-19)\n\Rightarrow I_p=4.06283984* 10^(-15)\ A$

Total current

$I=2.0235* 10^(-14)+4.06283984* 10^(-15)\n\Rightarrow I=2.429783984* 10^(-14)\ A$

The average current is $2.429783984* 10^(-14)\ A$

Those who support the alien theory believe that the ancient Egyptians where very primitive. What evidence about the ancient Egyptians suggests that claim is inaccurate?

Answers

Answer:

They had a well-developed agricultural system , they raised domesticated animals , they developed a writing system, and they used early forms of tools.

Final answer:

The claim that the ancient Egyptians were primitive and relied on aliens to build their monuments is inaccurate. Evidence from archaeology and history shows that the ancient Egyptians had advanced knowledge and skills in various fields. Their construction techniques and use of mathematics in building the pyramids are well-documented.

Explanation:

The claim that the ancient Egyptians were primitive and that their accomplishments, such as building the pyramids, were assisted by aliens is inaccurate. There is evidence from archaeology and history that ancient monuments were built by ancient people using their own ingenuity and capabilities. The ancient Egyptians had advanced knowledge in various fields including architecture, engineering, astronomy, and mathematics. For example, their construction techniques and use of mathematics in building the pyramids and other structures are well-documented.

Learn more about Ancient Egyptians

Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit. The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm. What is the width of the slit?

Answers

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, $\lambda=608\ nm=608* 10^(-9)\ m$

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

$y=(mD\lambda)/(a)$

where

a = width of the slit

$a=(mD\lambda)/(y)$

$a=(5* 0.885\ m* 608* 10^(-9)\ m)/(0.0161\ m)$

a = 0.000167 m

$a=1.67* 10^(-4)\ m$

a = 0.167 mm

So, the width of the slit is 0.167 mm. Hence, this is the required solution.

A child attaches a 0.6 kg toy to a 0.9 meter length of string and spins it around in uniform circular motion. Calculate the tension in the string if the period of rotation is 4.1 seconds.