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3 years ago

A mxiture of n2 and H2 has mole fraction of 0.4 and 0.6 respectively. Determine the density of the mixture at one bar and 0 c.

Physics

1 answer:

SOVA2 [1]3 years ago

3 0

Answer:

The density of the mixture is 0.55kg/m^3

Explanation:

P = 1bar = 100kN/m^2, T = 0°C = 273K, n = 0.4+0.6 = 1mole

PV = nRT

V = nRT/P = 1×8.314×273/100 = 22.70m^3

Mass of N2 = 0.4×28 = 11.2kg

Mass of H2 = 0.6×2 = 1.2kg

Mass of mixture = 11.2 + 1.2 = 12.4kg

Density of mixture = mass/volume = 12.4/22.7 = 0.55kg/m^3

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12,000/ 4000 = 3

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A rope of negligible mass passes over a uniform cylindrical pulley of 1.50kg massand 0.090m radius. The bearings of the pulley h

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Answer:

Explanation:

mass of pulley, m3 = 1.5 kg

Radius of pulley, R = 0.09 m

mass of monkey, m2 = 4.5 kg

mass of banana bunch, m1 = 3 kg

Let a is teh acceleration ans T1 and T2 be the tension in the rope.

The moment of inertia of the pulley

I = 0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²

According to Newton's second law

T1 - m1 g = m1 x a .... (1)

m2 g - T2 = m2 x a ..... (2)

(T2 - T1 ) x R = I x α

where, α is the angular acceleration

α = a / R

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T2 - T1 = 0.5 x m3 x a ..... (3)

from (1), (2) and (3)

$a = \frac{m_{2}-m_{1}}{m_{1}+m_{2}+\frac{m_{3}}{2}}\times g$

$a = \frac{4.5-3}{3+4.5+0.75}\times 9.8$

a = 1.78 m/s²

from equation (1)

T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N

from equation (2)

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3 years ago

During a race, four competitors of the same weight rode identical bicycles for 10 minutes. At 8 minutes, which bicycle was movin

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All the competitors will move with the same velocity.

Explanation:

Here, the situations for each competitor are identical. Thus, they will exert the same force and hence, their velocities at each instants will be identical.

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3 years ago

A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

Mashutka [201]

Answer:

$a = 0.53 m/s^2$

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

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$\omega = 2\pi ( \frac{20}{60})$

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so final tangential speed is given as

$v = r\omega$

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3 years ago

A car goes around a curve of radius r at a constant speed v. Then it goes around a curve ofradius 2r at speed 2v. What is the ce

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The centripetal force on the car as it goes around the second curve is twice that compared to the first.

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It is the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation.

The formula of Centripetal force is:

F(c) = (m* v^2) / r

Here,

At the first curve,

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The centripetal force on the car as it goes around the second curve is twice that compared to the first.

Learn more about centripetal force here:

<u>brainly.com/question/14317060</u>

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