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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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Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

balandron [24]

Answer with Explanation:

We are given that

Radius of solid core wire=r=2.28 mm= $2.28\times 10^{-3} m$

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

Radius of each strand of thin wire=r'=0.456 mm= $0.456\times 10^{-3} m$

Current density of each wire= $J=3750 A/m^2$

a.Area = $\pi r^2$

Where $\pi=3.14$

Using the formula

Cross section area of copper wire has solid core = $3.14\times (2.28\times 10^{-3})^2=16.3\times 10^{-6} m^2$

Current density = $J=\frac{I}{A}$

Using the formula

$3750=\frac{I}{16.3\times 10^{-6}}$

$I=3750\times 16.3\times 10^{-6}=0.061 A$

Total number of strands=19

Area of strand wire= $A'=19\times 3.14\times (0.456\times 10^{-3})^2=12.4\times 10^{-6} m^2$

$J'=\frac{I'}{A'}$

$3750=\frac{I'}{19\times 3.14(0.456\times 10^{-3})^2}$

$I'=3750\times 19\times 3.14(0.456\times 10^{-3})^2$

$I'=0.047 A$

b.Resistivity of copper wire= $\rho=1.69\times 10^{-8}\Omega-m$

Length of each wire =6.25 m

Resistance, R= $\frac{\rho l}{A}$

Using the formula

Resistance of solid core wire= $R=\frac{1.69\times 10^{-8}\times 6.25}{16.3\times 10^{-6}}=6.5\times 10^{-3}\Omega$

Resistance of strand wire= $R'=\frac{1.69\times 10^{-8}\times 6.25}{12.4\times 10^{-6}}=8.5\times 10^{-3}\Omega$

7 0

1 year ago

Para proteger un computador de sobrecargas eléctricas, Juan coloca un filamento delgado de cobre llamado fusible en su circuito,

Lynna [10]

Answer:

Los fusibles están diseñados de tal forma que estos se "rompen" o se funden, cuando la demanda eléctrica supera un dado valor (cuando demasiada electricidad pasa a través de el).

Una vez el filamento se rompe, la corriente ya no puede circular por el (podes pensar en esta situación como un cable roto, la electricidad no puede circular por este cable)

Entonces, al romperse el filamento, en caso de una sobrecarga eléctrica, el flujo de electricidad se corta, y de esta forma se protege al computador de posibles sobrecargas.

7 0

1 year ago

The position of a particle moving along the x-axis varies with time according to x(t) = 5.0t^2 − 4.0t^3 m. Find (a) the velocity

KengaRu [80]

<h2>Answer:</h2>

(a) v(t) = [10.0t - 12.0t²] m/s and a(t) = [10.0 - 24.0t ] m/s² respectively

(b) -28.0m/s and -38.0m/s² respectively

(c) 0.83s

(d) 0.83s

(e) x(t) = 1.1573 m [where t = 0.83s]

<h2>Explanation:</h2>

The position equation is given by;

x(t) = 5.0t² - 4.0t³ m --------------------(i)

(a) Since velocity is the time rate of change of position, the velocity, v(t), of the particle as a function of time is calculated by finding the derivative of equation (i) as follows;

v(t) = dx(t) / dt = $\frac{dx}{dt}$ = $\frac{d}{dt}$ [ 5.0t² - 4.0t³ ]

v(t) = 10.0t - 12.0t² --------------------------------(ii)

Therefore, the velocity as a function of time is v(t) = 10.0t - 12.0t² m/s

Also, since acceleration is the time rate of change of velocity, the acceleration, a(t), of the particle as a function of time is calculated by finding the derivative of equation (ii) as follows;

a(t) = dx(t) / dt = $\frac{dv}{dt}$ = $\frac{d}{dt}$ [ 10.0t - 12.0t² ]

a(t) = 10.0 - 24.0t --------------------------------(iii)

Therefore, the acceleration as a function of time is a(t) = 10.0 - 24.0t m/s²

(b) To calculate the velocity at time t = 2.0s, substitute the value of t = 2.0 into equation (ii) as follows;

=> v(t) = 10.0t - 12.0t²

=> v(2.0) = 10.0(2) - 12.0(2)²

=> v(2.0) = 20.0 - 48.0

=> v(2.0) = -28.0m/s

Also, to calculate the acceleration at time t = 2.0s, substitute the value of t = 2.0 into equation (iii) as follows;

=> a(t) = 10.0 - 24.0t

=> a(2.0) = 10.0 - 24.0(2)

=> a(2.0) = 10.0 - 48.0

=> a(2.0) = -38.0 m/s²

Therefore, the velocity and acceleration at t = 2.0s are respectively -28.0m/s and -38.0m/s²

(c) The time at which the position is maximum is the time at which there is no change in position or the change in position is zero. i.e dx / dt = 0. It also means the time at which the velocity is zero. (since velocity is dx / dt)

Therefore, substitute v = 0 into equation (ii) and solve for t as follows;

=> v(t) = 10.0t - 12.0t²

=> 0 = 10.0t - 12.0t²

=> 0 = ( 10.0 - 12.0t ) t

=> t = 0 or 10.0 - 12.0t = 0

=> t = 0 or 10.0 = 12.0t

=> t = 0 or t = 10.0 / 12.0

=> t = 0 or t = 0.83s

At t=0 or t = 0.83s, the position of the particle will be maximum.

To get the more correct answer, substitute t = 0 and t = 0.83 into equation (i) as follows;

<em>Substitute t = 0 into equation (i)</em>

x(t) = 5.0(0)² - 4.0(0)³ = 0

At t = 0; x = 0

<em>Substitute t = 0.83s into equation (i)</em>

x(t) = 5.0(0.83)² - 4.0(0.83)³

x(t) = 5.0(0.6889) - 4.0(0.5718)

x(t) = 3.4445 - 2.2872

x(t) = 1.1573 m

At t = 0.83; x = 1.1573 m

Therefore, since the value of x at t = 0.83s is 1.1573m is greater than the value of x at t = 0 which is 0m, then the time at which the position is at maximum is 0.83s

(d) The velocity will be zero when the position is maximum. That means that, it will take the same time calculated in (c) above for the velocity to be zero. i.e t = 0.83s

(e) The maximum position function is found when t = 0.83s as shown in (c) above;

Substitute t = 0.83s into equation (i)

x(t) = 5.0(0.83)² - 4.0(0.83)³

x(t) = 5.0(0.6889) - 4.0(0.5718)

x(t) = 3.4445 - 2.2872

x(t) = 1.1573 m [where t = 0.83s]

8 0

2 years ago

The speed of sound in dry air at 20 °C is 343.5 m s-1, and the frequency of the sound from the note C# above middle C on the pia

Nastasia [14]

Answer:

1.23917 m

0.14323 s

Explanation:

v = Speed of sound in dry air at 20 °C = 343.5 m/s

f = Frequency of note C# = 277.2 /s = 277.2 Hz

λ = Wavelength

$v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{343.5}{277.2}\\\Rightarrow \lambda=1.23917\ m$

Wavelength = 1.23917 m

Distance the wave needs to travel is 49.2 m

Time = Distance / Speed

$\text{Time}=\frac{49.2}{343.5}=0.14323\ s$

Time taken for the sound to travel across the concert hall is 0.14323 s

4 0

2 years ago

The chart shows the voltage of four electric currents.

Scorpion4ik [409]

Answer:

Current X has a lower potential difference than Current Y.

Explanation:

The table is as follows:

Current Volts (V)

W 9.0

X 1.5

Y 3.0

Z 4.5

There are two quantities represented in the table:

1) Current: the current is the rate of flow of electric charge in a circuit. It is given by

$I=\frac{q}{t}$

where q is the amount of charge that passes a given point of a circuit in a time t. It is measured in Ampere (A).

2) Potential difference: the potential difference is the difference in electric potential between two points of a circuit. The potential difference is responsible for "pushing" the electrons through the circuit and producing a current. It is measured in Volts (V).

From the table, we see that

Current X has a lower potential difference (1.5 Volts) than Current Y (3.0 Volts)

So the correct choice is

Current X has a lower potential difference than Current Y.

3 0

2 years ago

Read 2 more answers