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2 years ago

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Can pockets of vacuum persist in an ideal gas? Assume that a room is filled with air at 20∘C and that somehow a small spherical

region of radius 1 cm within the room becomes devoid of air molecules. Estimate how long it will take for air to refill this region of vacuum. Assume the atomic mass of air is 29 u.

1 answer:

Ilya [14]2 years ago

3 0

Answer:

time taken is 20 μs

Explanation:

given data

temperature = 20°C = 293 K

radius = 1 cm

atomic mass of air = 29 u

to find out

how long it will take for air to refill

solution

we find here rms velocity of air particle that is

$\frac{1}{2}mv^2 = \frac{3}{2}RT$

here m is mass and t is temperature and v is speed and R is ideal gas constant i.e. 8.3145 (kg·m²/s²)/K·mol

v = $\sqrt{\frac{3RT}{M} }$ ............................1

v = $\sqrt{\frac{3(8.314)293}{29*10{-3}kg} }$

v = 501.99 m/s

so now for cover 1 cm

time taken by air

time take = $\frac{r}{v}$

time taken = $\frac{1*10^{-2}m}{501.99}$

time taken = 19.92 × $10^{6}$ s = 20μs

so time taken is 20 μs

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You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle

charle [14.2K]

Answer:

a= $5.54m/s^{2}$

Explanation:

The net force, $F_{net}$ of the box is expressed as a product of acceleration and mass hence

$F_{net}=ma$ where m is mass and a is acceleration

Making a the subject, a= $\frac {F_{net}}{m}$

From the attached sketch,

∑ $F_{net}=Fcos\theta-F_{f}$ where $F_{f}$ is frictional force and $\theta$ is horizontal angle

Substituting ∑ $F_{net}$ as $F_{net}$ in the equation where we made a the subject

a= $\frac {Fcos\theta-F_{f}}{m}$

Since we’re given the value of F as 240N, $F_{f}$ as 41.5N, $\theta$ as $30^{o}$ and mass m as 30kg

a= $\frac {240cos30-41.5}{30.0}=\frac {166.346}{30.0}=5.54m/s^{2}$

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2 years ago

How high above the earth's surface is g reduced to 8.80m/^2?

Sladkaya [172]

Gravity changes as the altitude change.<span> The gravitational force is proportional to 1/R2, where R is the distance from the center of the Earth the radius of earth where gravity is 9.8 m/s^2 is 6400 km this will serve as the zero mark.

g1/(g2) = R2^2/(R1)^2

so we set the constant values to R1 and the unknown distance as x

(9.8)/(8.80) = (6400-x)2/(6400)^2

solving for x we will get

x = 345.85 km above the earths surface
</span>

<span>Hope my answer would be a great help for you.
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2 years ago

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he drawing shows two perpendicular, long, straight wires, both of which lie in the plane of the paper. The current in each of th

AleksandrR [38]

Answer:

The magnitudes of the net magnetic fields at points A and B is 2.66 x $10^{-6}$ T

Explanation:

Given information :

The current of each wires, I = 4.7 A

dH = 0.19 m

dV = 0.41 m

The magnetic of straight-current wire :

B= μ $_{0}$ I/2πr

where

B = magnetic field (T)

μ $_{0}$ = 1.26 x $10^{-6}$ (N/ $A^{2}$ )

I = Current (A)

r = radius (m)

the magnetic field at points A and B is the same because both of wires have the same distance. Based on the right-hand rule, the net magnetic field of A and B is canceled each other (or substracted). Thus,

BH = μ $_{0}$ I/2πr

= (1.26 x $10^{-6}$ )(4.7)/(2π)(0.19)

= 4.96 x $10^{-6}$ T

BV = μ $_{0}$ I/2πr

= (1.26 x $10^{-6}$ )(4.7)/(2π)(0.41)

= 2.3 x $10^{-6}$ T

hence,

the net magnetic field = BH - BV

= 4.96 x $10^{-6}$ - 2.3 x $10^{-6}$

= 2.66 x $10^{-6}$ T

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1 year ago

Compared to the resistivity of a 0.4-meter length of 1-millimeter-diameter copper wire at 0 degrees Celsius, the resistivity of

Mazyrski [523]

Answer:

Resistivity of both wires are same

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Diameter, $d_1=1mm$

Radius, $r_1=\frac{d_1}{2}=\frac{1}{2}mm=0.5\times 10^{-3} m$

$1mm=10^{-3} m$

$l_2=0.8 m$

$d_2=1mm$

$r_2=0.5\times 10^{-3} m$

Temperature in each case is same.

Area of each wire, $A_1=A_2=A=\pi r^2=\pi (0.5\times 10^{-3})^2m^2$

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Resistivity of material depends upon the temperature and material by which it is made.

It does not depends upon the length of object.

Therefore, the resistivity of both wires of different length are same.

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