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AleksAgata [21]
3 years ago
10

How is the formula for finding the speed of a wave like the formula for finding the speed of a person running a race ?

Physics
1 answer:
Harman [31]3 years ago
3 0
Wave has distance called wavelength and time is same for both
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3 years ago
A. How many atoms of helium gas fill a spherical balloon of diameter 29.6 cm at 19.0°C and 1.00 atm? b. What is the average kine
Korolek [52]

Answer:

a) 3.39 × 10²³ atoms

b) 6.04 × 10⁻²¹ J

c) 1349.35 m/s

Explanation:

Given:

Diameter of the balloon, d = 29.6 cm = 0.296 m

Temperature, T = 19.0° C = 19 + 273 = 292 K

Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa

Volume of the balloon = \frac{4}{3}\pi(\frac{d}{2})^3

or

Volume of the balloon = \frac{4}{3}\pi(\frac{0.296}{2})^3

or

Volume of the balloon, V = 0.0135 m³

Now,

From the relation,

PV = nRT

where,

n is the number of moles

R is the ideal gas constant = 8.314  kg⋅m²/s²⋅K⋅mol

on substituting the respective values, we get

1.013 × 10⁵ × 0.0135 = n × 8.314 × 292

or

n = 0.563

1 mol = 6.022 × 10²³ atoms

Thus,

0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms

b) Average kinetic energy = \frac{3}{2}\times K_BT

where,

Boltzmann constant, K_B=1.3807\times10^{-23}J/K

Average kinetic energy = \frac{3}{2}\times1.3807\times10^{-23}\times292

or

Average kinetic energy = 6.04 × 10⁻²¹ J

c) rms speed = \frac{3RT}{m}

where, m is the molar mass of the Helium = 0.004 Kg

or

rms speed = \frac{3\times8.314\times292}{0.004}

or

rms speed = 1349.35 m/s

5 0
3 years ago
Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/
Mama L [17]

Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.

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Which term is applied to an object through which light passes?
Elena L [17]

Answer:

D

Explanation:

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