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geniusboy
10 days ago
5

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.
Physics
1 answer:
kicyunya [1K]10 days ago
3 0

Answer:

The time required is 20 μs

Explanation:

Here is the data provided:

temperature = 20°C  = 293 K

radius = 1 cm

atomic mass of air = 29 u

To determine

the duration for air to refill the vacuum space

solution:

We calculate the root mean square velocity of air particles. This can be expressed as:

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

where m indicates mass, t is temperature, v is speed, and R is the ideal gas constant, which is approximately 8.3145 (kg·m²/s²)/K·mol.

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

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

Resulting in v = 501.99 m/s.

<pNow, to cover the distance of 1 cm,<pThe duration needed for air is calculated as:

time taken = \frac{r}{v}

which gives us:

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

so, time taken = 19.92 × 10^{6} seconds = 20μs.

Thus, the required time is 20 μs.

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An object is attached to a hanging unstretched ideal and massless spring and slowly lowered to its equilibrium position, a dista
Sav [1105]

Answer:

        h = 12.8 cm

Explanation:

The initial parameters are as follows:

distance = 6.4 cm

  • when the object descends, its weight matches the spring's force

        weight = spring force

         mg = ky... equation 1

  • potential energy stored in a stretched spring = work done by the spring

        mgh = 0.5 x k x h^{2}....equation 2

  • Substituting from equation 1 into equation 2

                kyh =  0.5 x k x h^{2}

                y =  0.5 x h

                2y = h

  • where y is 6.4, yielding the maximum elongation as

          h = 2 x 6.4 = 12.8 cm

6 0
5 days ago
A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b
Softa [913]

Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b,   noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a.  Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.

8 0
10 days ago
Consider the uniform electric field \vec{E} =(4000~\hat{j}+3000~\hat{k})~\text{N/C} ​E ​⃗ ​​ =(4000 ​j ​^ ​​ +3000 ​k ​^ ​​ ) N/
kicyunya [1025]

Answer:

Electric flux is calculated as \phi=31562.63\ Nm^2/C

Explanation:

We start with the given parameters:

The electric field impacting the circular surface is E=(4000j+3000k)\ N/C

Our objective is to ascertain the electric flux passing through a circular region with a radius of 1.83 m situated in the xy-plane. The area vector is oriented in the z direction. The formula for electric flux is expressed as:

\phi=E{\cdot}A

\phi=(4000j+3000k){\cdot}Ak

Applying properties of the dot product, we calculate the electric flux as:

\phi=3000\times Ak

\phi=3000\times \pi (1.83)^2

\phi=31562.63\ Nm^2/C

Consequently, the electric flux for the circular area is \phi=31562.63\ Nm^2/C. Thus, this represents the required answer.

4 0
1 day ago
A champion athlete can produce one horsepower (746 W) for a short period of time. The number of 16-cm-high steps a 70-kg athlete
Sav [1105]

Answer:

407 steps

Explanation:

Based on the question,

P = mgh/t........... Equation 1

Where P stands for power, m is mass, g denotes gravity, h is height, and t represents time.

Rearranging the equation to solve for h, we have:

h = Pt/mg............. Equation 2

Providing values: P = 746 W, t = 1 minute = 60 seconds, m = 70 kg.

Given constant: g = 9.8 m/s²

By substituting into equation 2

h = 746(60)/(70×9.8)

h = 44760/686

h = 65.25 m

h = 6525 cm

Calculating number of steps: 6525/16

The resulting number of steps = 407 steps

6 0
9 days ago
One consequence of Einstein's theory of special relativity is that mass is a form of energy. This mass-energy relationship is pe
ValentinkaMS [1149]

Answer:

The energy unit is expressed as kg-m/s or Joules.

Explanation:

The relationship between mass and energy in physics is represented by:

E=mc^2

Where

m denotes the mass of the object

c signifies the speed of light

In the SI system, mass is measured in kilograms and the speed of light in m/s. Therefore, energy is defined in kg-m/s, which is equal to Joules.

Thus, the appropriate SI unit for energy is kg-m/s or Joules. This concludes the explanation.

6 0
8 days ago
Read 2 more answers
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