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Aliun [14]
3 years ago
13

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

tic energy of the helium atoms?
c. What is the rms speed of the helium atoms?
Physics
1 answer:
Korolek [52]3 years ago
5 0

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

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In order from shortest to longest frequency, the electromagnetic waves can be ranked :

1. X-rays

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A 10kg sphere hits a stationary 8kg sphere. After the collision the 8kg sphere moves off in the positive direction at 4m/s. If t
Bumek [7]

M = mass of the first sphere = 10 kg

m = mass of the second sphere = 8 kg

V = initial velocity of the first sphere before collision = 10 m/s

v = initial velocity of the second sphere before collision = 0 m/s

V' = final velocity of the first sphere after collision = ?

v' = final velocity of the second sphere after collision = 4 m/s

using conservation of momentum

M V + m v = M V' + m v'

(10) (10) + (8) (0) = (10) V' + (8) (4)

100 = (10) V' + 32

(10) V' = 68

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3 years ago
Three ropes A, B and C are tied together in one single knot K.
arsen [322]

The tension in the rope B is determined as 10.9 N.

<h3>Vertical angle of cable B</h3>

tanθ = (6 - 4)/(5 - 0)

tan θ = (2)/(5)

tan θ = 0.4

θ = arc tan(0.4) = 21.8 ⁰

<h3>Angle between B and C</h3>

θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰

Apply cosine rule to determine the tension in rope B;

A² = B² + C² - 2BC(cos A)

B = C

A² = B² + B² - (2B²)(cos A)

A² = 2B² - 2B²(cos 43.6)

A² = 0.55B²

B² = A²/0.55

B² = 65.3/0.55

B² = 118.73

B = √(118.73)

B = 10.9 N

Thus, the tension in the rope B is determined as 10.9 N.

Learn more about tension here: brainly.com/question/24994188

#SPJ1

8 0
2 years ago
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PLEASE HELP ASAP 30 POINTS !!!!!!
crimeas [40]

Answer:

<h3><u>If the mass of both of the objects is doubled, then the force of gravity between them is quadrupled; and so on. Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces.</u></h3>
7 0
3 years ago
A man pushes a box along a flat, frictionless surface using a force of 500 N. The box was moved a distance of 2.5 m. The actual
andrey2020 [161]

Answer:

Workdone = 1250Nm

Explanation:

<u>Given the following data;</u>

Force, F = 500N

Distance, d = 2.5m

Workdone is given by the formula;

Workdone = force * distance

Substituting into the equation, we have

Workdone = 500 * 2.5  

Workdone = 1250Nm

Therefore, the actual work done by the worker is 1250 Newton-meter.

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