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3 years ago

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What is the root mean square speed of an As4 particle as it is sublimed? (Assume at high temperatures arsenic acts like an ideal

gas; the Boltzmann constant, kB, can be approximated as 1.4 x 10-23 J∙K-1) and the kinetic energy of the gas is equal to 3/2kB T

1 answer:

Norma-Jean [14]3 years ago

5 0

Answer: $v_{rms} =$ 273m/s

Explanation: <u>Root</u> <u>Mean</u> <u>Square</u> <u>Speed</u> of an atom or molecule is the speed of a particle in a gas. It is the average speed a particle in a gas can have.

It can be calculated:

$\frac{1}{2}mv^{2} = \frac{3}{2} k_{B}T$

$v^{2} = \frac{3k_{B}T}{m}$

$v = \sqrt{\frac{3k_{B}T}{m}}$

m is mass of one atom or molecule in kg.

An atom of Arsenic sublimes at 614°C. Converting to Kelvin:

T = 614 + 273 = 887K

Molecular mass of As4 is approximately 0.3kg.

$m = \frac{0.3}{6.10^{23}}$

$m=5.10^{-25}$ kg

Calculating Root mean square speed :

$v = \sqrt{\frac{3*1.4.10^{-23}*887}{5.10^{-25}}$

$v = \sqrt{745.08.10^{2}$

v = 273m/s

<u>The root mean square speed of As4 is approximately </u><u>273m/s</u><u>.</u>

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A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist

Paha777 [63]

Answer:

Part a)

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

$\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

$\frac{1}{2}mv^2 = mgh'$

now we know that

$I = \frac{2}{5}mr^2$

$\omega = \frac{v}{r}$

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$

$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh$

$\frac{7}{10}mv^2 = mgh$

$\frac{1}{2}mv^2 = \frac{10}{14}mgh$

so the height on the smooth side is given as

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

6 0

4 years ago

Read 2 more answers

a container of water is knocked off a 10.0 meter high ledge with a horizontal velocity of 1.00 meters/second. calculate the time

Evgen [1.6K]

Answer:

1.43 s

Explanation:

The time it takes for the container to reach the ground is determined only by the vertical motion of the container, which is a free-fall motion, so a uniformly accelerated motion with a constant acceleration of g=9.8 m/s^2 towards the ground.

The vertical distance covered by an object in free fall is given by

$S=ut + \frac{1}{2}at^2$

where

u = 0 is the initial vertical speed

t is the time

a= g = 9.8 m/s^2 is the acceleration

since u=0, it can be rewritten as

$S=\frac{1}{2}gt^2$

And substituting S=10.0 m, we can solve for t, to find the duration of the fall:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(10.0 m)}{9.8 m/s^2}}=1.43 s$

3 0

4 years ago

Assuming 84.0% efficiency for the conversion of electrical power by the motor, what current must the 13.0-V batteries of a 716 k

tino4ka555 [31]

Answer:

$\mathbf{ current(I) =1766.67 \ A}$

Explanation:

Given that:

The air resistance and friction = 700 N

The gravity caused force = 716 × 9.8 = 7016.8

Total force = (7016.8 + 700) N

Total force = 7716.8 N

∴

$13 \times current(I) \times 0.84 = \dfrac{7716.8 \times 300}{2 \times 60}$

$current(I) \times 10.92= 19292$

$current(I) = \dfrac{19292}{10.92}$

$\mathbf{ current(I) =1766.67 \ A}$

8 0

3 years ago

What is the volume V of a sample of 4.00 mol of copper? The atomic mass of copper (Cu) is 63.5 g/mol, and the density of copper

rusak2 [61]

Answer : The volume of a sample of 4.00 mol of copper is $28.5cm^3$

Explanation :

First we have to calculate the mass of copper.

$\text{ Mass of copper}=\text{ Moles of copper}\times \text{ Molar mass of copper}$

$\text{ Mass of copper}=(4.00moles)\times (63.5g/mole)=254g$

Now we have to calculate the volume of copper.

Formula used :

$Density=\frac{Mass}{Volume}$

Now put all the given values in this formula, we get:

$8.92\times 10^3kg/m^3=\frac{254g}{Volume}$

$Volume=\frac{254g}{8.92\times 10^3kg/m^3}=2.85\times 10^{-2}L=2.85\times 10^{-2}\times 10^3cm^3=28.5cm^3$

Conversion used :

$1kg/m^3=1g/L\\\\1L=10^3cm^3$

Therefore, the volume of a sample of 4.00 mol of copper is $28.5cm^3$

7 0

3 years ago

Read 2 more answers

Can someone help me plzzz..<br>whoever answers the best will be marked as brainliest.....

IrinaK [193]

Answer:

1) 3 applications of pressure in daily life are :-

● The area of sharp edge of knife, scissor or handsaws are much less then blunt edge. So, for same total force pressure is more for sharp edges than the blunt one. Hence sharp knife, scissors etc, cuts easily than a blunt one.

●Broad handles in bags and suitcases are provided for the comfort. Broad handles have large area. So, the pressure exerted on hands and shoulders would be small while carrying the bags and the suitcases.

●Trucks carrying heavy loads have more than four tyres. More tyres in case of trucks increase the area of contact with the road. This results in reduced pressure on the tyres.

2) Area of the surface which is on ground = 1.5×1

= 1.5m^2

Mass of the block = 300kg

Force applied by the block = Mass × g = 300×10

= 3000N (where g = acceleration due to gravity )

Pressure = Force applied / Area of the surface

= 3000N / 1.5m^2

= 2000 Pa

3)

a) The above experiment signifies that more the area of the surface of an object , less the pressure an object applies.

b) B exerts the minimum pressure because the area of its surface to ground is greater than others & as it has more area of surface , it exerts less pressure. ( area is inversely proportional to pressure )

c) D exerts the maximum pressure because the area of its surface to ground is lesser than others & as it has less area of surface , it exerts more pressure. ( area is inversely proportional to pressure )

d) It depend upon the way an object is kept on ground. If an object is kept in such a way dat the area of the surface to the ground is more , then pressure will be least exerted .If an object is kept in such a way dat the area of the surface to the ground is less, then pressure will be exerted more .

e) Do it yourself . only i will suggest that make the tip of the cone ( which is to the ground ) more narrower.

6 0

3 years ago