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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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A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of What is the

Alborosie

Answer:

<em>0.45 mm</em>

Explanation:

The complete question is

a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

A) 0.45 mm

B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

Current in the fuse is 1.0 A

Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = $\frac{\pi d^{2} }{4}$

where d is the diameter of the wire

1.613 x 10^-7 = $\frac{3.142* d^{2} }{4}$

6.448 x 10^-7 = 3.142 x $d^{2}$

$d^{2}$ = $\sqrt{ 2.05*10^-7}$

d = 4.5 x 10^-4 m = <em>0.45 mm</em>

8 0

2 years ago

Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is

ad-work [718]

Answer:

$T_{out}=27.0000077$ ºC

Explanation:

First, let's write the energy balance over the duct:

$H_{out}=H_{in}+Q$

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

$m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q$

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

$m*Cp(T_{out}-T_{in})=Q$

So, let's isolate $T_{out}$ :

$T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}$

The Cp of the air at 27ºC is 1007 $\frac{J}{kgK}$ (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are $T_{out}$ and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

$P=2*H+2*A=2*0.004m+2*0.016m=0.04m$

Surface area:

$A=P*L=0.04m*1m=0.04m^2$

Then, the heat Q is:

$600\frac{W}{m^2} *0.04m^2=24W$

Finally, find the exit temperature:

$T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077$

$T_{out}$ =27.0000077 ºC

The temperature change so little because:

The mass flow is so big compared to the heat flux.
The transfer area is so little, a bigger length would be required.

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2 years ago

In the winter sport of bobsledding, athletes push their sled along a horizontal ice surface and then hop on the sled as it start

Hitman42 [59]

Answer:

$v_f = 16.6 m/s$

Explanation:

As we know by force equation that force along the inclined planed due to gravity is given as

$F_g = mg sin\theta$

so the acceleration due to gravity along the plane is given as

$a = \frac{F_g}{m}$

now we have

$a = g sin\theta$

$a = (9.81 sin4.0)$

$a = 0.68 m/s^2$

now we know that

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 9.2^2 = 2(0.68)(140)$

$v_f = 16.6 m/s$

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2 years ago

The drawing shows an object attached to an ideal spring, which is hanging from the ceiling. The unstrained length of the spring

Andrew [12]

Image is missing so I have attached it.

Also, the options are missing and it is;

A) KE is has a maximum value at position 3. EPE has a maximum value at position 2. GPE has a maximum value at position 1.

B) KE is has a maximum value at position 1. EPE has a maximum value at position 2. GPE has a maximum value at position 3.

C) KE is has a maximum value at position 2. EPE has a maximum value at position 3. GPE has a maximum value at position 1.

D) KE is has a maximum value at position 1. EPE has a maximum value at position 3. GPE has a maximum value at position 2.

E) KE is has a maximum value at position 2. EPE has a maximum value at position 1. GPE has a maximum value at position 3.

Answer:

Option C is the correct answer which says; KE is has a maximum value at position 2. EPE has a maximum value at position 3. GPE has a maximum value at position 1.

Explanation:

If an object vibrates about its mean position, under the influence of a restoring force, such that restoring force is directly proportional to the displacement from the mean position, the motion of the object is called simple harmonic motion. During Simple harmonic motion, the sum of Kinetic and potential energy remains constant.

Now, Looking at the diagram, Kinetic Energy (KE) is maximum at position 2.

Looking at the options, only C and E agree with this.

Thus, our answer is either option C or E.

However, Option E is not going to be right because it says that GPE is at a maximum at position 3, which is not true as the maximum GPE will occur at position 1.

Thus,

Option C fulfills that and therefore will be the correct answer.

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2 years ago

A child sets off the firecracker at a distance of 100 m from the family house. what is the sound intensity β100 at the house?

KATRIN_1 [288]

To solve this problem, we use the formula:

I100 / I1 = [P / 4π(100m)^2] / [P / 4π(1m)^2]

I100 / I1 = 1 / 100^2

I100 / I1 = 10^-4

Therefore the change in intensity from 1m to 100m in decibels is:

B100 – B1 = 10 log(10^-4) dB = -40 dB

So the intensity at 100m is calculated as:

B100 = B1 – 40 dB = 140 dB – 40 dB = 100 dB

Answer:

100 dB

6 0

2 years ago

Read 2 more answers