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3 years ago

14

Equal masses of he and ne are placed in a sealed container. What is the partial pressure of he if the total pressure in the cont

ainer is 6 atm?

1 answer:

EastWind [94]3 years ago

6 0

Answer:

6 atm.

Explanation:

Let the mass of both be m

Then moles of He = m/ 4

Moles of Ne = m/ 20

mole fraction of He = Moles of He/ Total moles = m/4/ (m/4 + m/20) = 0.25 m/0.3m = 0.83

Pressure of He = Mole fraction×total pressure = 0.83 × 6 atm = 5 atm

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A hamster of mass 0.139 kg gets onto his 20.8−cm-diameter exercise wheel and runs along inside the wheel for 0.823 s until its f

Anna11 [10]

Explanation:

Given that,

Mass of the hamster, m = 0.139 kg

Diameter of the wheel, d = 20.8 cm

Radius, r = 10.4 m

Frequency of the wheel, f = 1 Hz

Time, t = 0.823 s

(a) There exist a relationship between the tangential and angular acceleration of the wheel. It is given by :

$a=\alpha \times r$

Since, $\alpha =\dfrac{\omega}{t}$

$a=\dfrac{\omega}{t} \times r$

$a=\dfrac{2\pi f}{t} \times r$

$a=\dfrac{2\pi \times 1}{ 0.823} \times 10.4\times 10^{-2}$

$a=0.793\ m/s^2$

(b) In radial direction, applying the second law of motion as :

$N-mg=ma$

a is the radial acceleration, $a=\dfrac{v^2}{r}$

$N=mg+ma$

$N=mg+m(\dfrac{v^2}{r})$

$N=mg+m(\dfrac{(r\omega)^2}{r})$

$N=mg+m\omega^2 r$

$N=m(g+\omega^2 r)$

$N=m(g+(2\pi f)^2 r)$

$N=0.139\times (9.8+(2\pi 1)^2\times 10.4\times 10^{-2})$

$N=1.93\ N$

Hence, this is the required solution.

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4 years ago

Properties of most medals include

Aleonysh [2.5K]

Answer:

Metals are lustrous, malleable, ductile, good conductors of heat and electricity. Other properties include: State: Metals are solids at room temperature with the exception of mercury, which is liquid at room temperature (Gallium is liquid on hot days).

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3 years ago

1. How much work is done by a person if he lifts a load of 500 kg upto a height of 2m?

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Explanation:

if we take g=9.8m/s^2

F=mg

m=500

F=500×9.8=4900N

h=2

so w=mgh

w=4900×2

=9800j

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3 years ago

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Answer:

wwadaww

Explanation:

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3 years ago

Read 2 more answers

When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the

Taya2010 [7]

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

$\theta = 54.6^{\circ}$

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = $\sqrt{\frac{gx}{sin2\theta}}$

v = $\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}$

v = $\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}$

v = $\sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s$

Now,

The angular velocity can be calculated as:

$v = \omega R$

Thus

$\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s$

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3 years ago