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wel
2 years ago
6

Approximately 1.000 g each of four gasses H2, Ne, Ar, and Kr are placed in a sealed container all under1.5 atm of pressure. Assu

ming ideal behavior, determine the partial pressure of the H2 and Ne
Physics
1 answer:
Vera_Pavlovna [14]2 years ago
7 0

Answer:

The partial pressure of H2 is 0.375 atm

The partial pressure of Ne is also 0.375 atm

Explanation:

Mass of H2 = 1 g

Mass of Ne = 1 g

Mass of Ar = 1 g

Mass of Kr = 1 g

Total mass of gas mixture = 1 + 1 + 1 + 1 = 4 g

Pressure of sealed container = 1.5 atm

Partial pressure of H2 = (mass of H2/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm

Partial pressure of Ne = (mass of Ne/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm

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A squirrel in a tree drops an acorn. how long does it take the acorn to fall 20 feet?
mart [117]

We use the equation of motion,

S= ut+\frac{1}{2}at^{2}

Here, S is the height, u is initial velocity and a is acceleration.

Given, S = 20 \ ft S = 20 \ ft = 20 \times\frac{1 \ m}{3.2808399 ft}  = 6.096 \ m

As  acorn falls from tree, therefore we take the value of a = 9.8 \ m/s^2 and initial velocity u = 0.

Substituting these values in equation of motion,

6.096 \ m = 0 \times t +\frac{1}{2} \times 9.8 m/s^2 (t)^2 \\\\\ t = 1.12 \ s

Thus, the time taken by the acorn to fall 20  feet ( 6.096 m ) is 1.12 s.

5 0
2 years ago
What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from
Strike441 [17]

To develop this problem we will apply the concepts related to the Doppler effect. The frequency of sound perceive by observer changes from source emitting the sound. The frequency received by observer f_{obs} is more than the frequency emitted by the source. The expression to find the frequency received by the person is,

f_{obs} = f_s (\frac{v_w}{v_w-v_s})

f_s= Frequency of the source

v_w= Speed of sound

v_s= Speed of source

The velocity of the ambulance is

v_s = 119km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

v_s = 30.55m/s

Replacing at the expression to frequency of observer we have,

f_{obs} = 800Hz(\frac{345m/s}{345m/s-30.55m/s})

f_{obs} = 878Hz

Therefore the frequency receive by observer is 878Hz

8 0
2 years ago
Mt. Asama, Japan, is an active volcano complex. In 2009, an eruption threw solid volcanic rocks that landed far from the crater.
solong [7]

Answer:u=97.41m/s

Explanation:

Given

inclination \theta =58.7^{\circ}C

Horizontal distance travel by Particle d=1200 m

Vertical height h=780 m

Let u be the initial velocity

calculating vertical distance

y=u\sin \theta +\frac{at^2}{2}

y=u\sin \theta t-\frac{gt^2}{2}-------1

Calculating horizontal distance

x=u\cos \theta \times t+0

t=\frac{x}{u\cos \theta }

put value of t in equation 1

y=u\sin \theta \times \frac{x}{u\cos \theta }-\frac{g}{2}\times (\frac{x}{u\cos \theta })^2

y=x\tan \theta -\frac{gx^2}{2u^2\cos ^2\theta }

\frac{gx^2}{2u^2\cos ^2\theta }=x\tan \theta -y

u^2=\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}

u=\sqrt{\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}}

at y=-780\ m\ x=1200 m

u^2=\frac{18.154}{2753.65}\times 1200^2

u=97.41 m/s

6 0
2 years ago
A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is its radius? (The value of μ0 is 4π x 1
nevsk [136]

Answer:

A. 2.2*10^-2m

Explanation:

Using

Area = length x L/ uo xN²

So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*

3000²)

A = 17.5*10^-3/ 1.13*10^-5

= 15.5*10^-2m²

Area= π r ²

15.5E-2/3.142 = r²

2.2*10^2m

Explanation:

5 0
2 years ago
the arm of a crane is 15.0 m long and makes an angle of 70.0 degrees with the horizontal. Assume that the maximum load for the c
Gnoma [55]

To find max torque you need component of force that is perpendicular to the direction of vector r. For second part just set the equation above equal to 2308 and solve for Fmax. Only difference is the angle

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