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3 years ago

What is meant by Compression and Rarefaction of a longitudinal wave?

1 answer:

6 0

However instead of crests and troughs, longitudinal waves have compressions and rarefactions. Compression. A compression is a region in a longitudinal wave where the particles are closest together. Rarefaction. A rarefaction is a region in a longitudinal wave where the particles are furthest apart.

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B. On a separate sheet of paper, describe the different ways of generating electric power.

Afina-wow [57]

Answer:

These all different sources of energy add to the store of electrical power that is then sent out to different locations via high powered lines. It is the energy from the sun that is harnessed using a range of technologies such as solar heating, solar architecture, photovoltaics, and artificial photosynthesis.

Hope it helps PLS MARK ME AS BRAINLIST I BEG YOU thanks :)

4 0

2 years ago

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

8 0

4 years ago

19. A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to

Aleks [24]

Recall this gas law:

$\frac{P₁V₁}{T₁}$ = $\frac{P₂V}{T₂}$

P₁ and P₂ are the initial and final pressures.

V₁ and V₂ are the initial and final volumes.

T₁ and T₂ are the initial and final temperatures.

Given values:

P₁ = 475kPa

V₁ = 4m³, V₂ = 6.5m³

T₁ = 290K, T₂ = 277K

Substitute the terms in the equation with the given values and solve for Pf:

$\frac{475*4}{290} = \frac{P₂*6.5}{277}$

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3 years ago

A metabolic waste of algae that can be recycled for use in cellular respiration is

m_a_m_a [10]

Answer:oxygen

Explanation:algae are plants that lives in aquatic habitat,with a few of them occurring in land. Algae have chlorophyll and as a result are autotrophic in nutrition. Algae uses carbon dioxide as a raw material for photosynthesis which is the process where they produce food. They also give off oxygen which results from the splitting of water by light.

This oxygen given off is used by organisms for cellular respiration.the mitochondria is the organelle responsible it's utilization in respiration and carbon dioxide is given off.oxygen serves as an electron acceptor in the energy producing process in the mitochondria. It is an important gas for aerobic respiration.

5 0

3 years ago

What factors affect the speed of water waves

Aneli [31]

Hey there,

Your question states: What factors affect the speed of water waves
Let's get one thing out the way, (wavelength) does $NOT$ affect the the speed of water. If anything, it would be how high the wavelength's are. The higher the wavelengths are, the more that it would affect the speed, because there very high, but if it were to go longer on the width side, that would increase the speed, but that's not the case. Your correct answer would be (higher wavelength).

Hope this really helps you.

6 0

3 years ago