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Gennadij [26K]
3 years ago
7

If the partial pressures of the three gases in the tank are 38.39 atm of O2, 3.38 atm of He, and 25.33 atm of N2, what is the to

tal partial pressure in the tank
Physics
1 answer:
Tasya [4]3 years ago
4 0

Answer:

67.1 atm

Explanation:

From Dalton's law of Partial Pressures, the total partial pressure equals the sum of the individual partial pressures.

P = P₁ + P₂ + P₃

P₁ = 38.39 atm of O₂

P₂ = 3.38 atm of He

P₃ = 25.33 atm of N₂

So, P = P₁ + P₂ + P₃ = 38.39 atm + 3.38 atm + 25.33 atm = 67.1 atm

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3 years ago
Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . The velo
yanalaym [24]

Answer:

Plane will 741.6959 m apart after 1.7 hour                    

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is   = 102^{\circ}-65.3^{\circ}=36.7^{\circ}

Speed of first plane  = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, r^2 representing the distance between the planes, we see that:

r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295

r = 741.6959 m

3 0
3 years ago
The main illustration in the video shows the life track of a one-solar mass star. Each point along this track represents _______
REY [17]

Each point along the track of one solar mass star represents the star's surface temperature and luminosity at one time.

<h3>What is the one-solar mass star?</h3>

A star having a mass equal to the mass of the Sun is called a one-solar mass star.

Its life track shows the luminous intensity as well as the surface temperature.

Learn more about one-solar mass star.

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2 years ago
35 miles North 6 feet down 136 MB 85 km SE
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3 years ago
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A closely wound search coil has an area of 3.21 cm2, 120 turns, and a resistance of 58.7 O. It is connected to a charge-measurin
Alexxx [7]

Answer:

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

\epsilon = \frac{NBA}{t}

Where,

\epsilon  =electromotive force

N = Number of loops

B = Magnetic field

A = Area

t= Time

For Ohm's law we now that,

V = IR

Where,

I = Current

R = Resistance

V = Voltage (Same that the electromotive force at this case)

In this system we have that the resistance in series of coil and charge measuring device is given by,

R = R_c + R_d

And that the current can be expressed as function of charge and time, then

I = \frac{q}{t}

Equation Faraday's law and Ohm's law we have,

V = \epsilon

IR = \frac{NBA}{t}

(\frac{q}{t})(R_c+R_d) = \frac{NBA}{t}

Re-arrange for Magnetic Field B, we have

B = \frac{q(R_c+R_d)}{NA}

Our values are given as,

R_c = 58.7\Omega

R_d = 45.5\Omega

N = 120

q = 3.53*10^{-5}C

A = 3.21cm^2 = 3.21*10^{-4}m^2

Replacing,

B = \frac{(3.53*10^{-5})(58.7+45.5)}{120*3.21*10^{-4}}

B = 0.095T

Therefore the magnetic field in the System is 0.095T

3 0
3 years ago
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