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s344n2d4d5 [400]
3 years ago
13

Boltzmann’s constant is 1.38066 × 10−23 J/K, and the universal gas constant is 8.31451 J/K · mol.

Physics
1 answer:
djyliett [7]3 years ago
4 0

Answer:

4.95\cdot 10^{-21} J

Explanation:

First of all, let's convert everything into SI units:

n = 2.4 mol (number of gas moles)

p=11 atm = 1.11\cdot 10^6 Pa (gas pressure)

V=4.3 L=4.3\cdot 10^{-3} m^3 (gas volume)

R = 8.31451 J/K · mol (gas constant)

The ideal gas equation states that

pV=nRT

Solving for T, we find the gas temperature

T=\frac{pV}{nR}=\frac{(1.11\cdot 10^6)(4.3\cdot 10^{-3})}{(2.4)(8.31451)}=239.2 K

And now we can find the average kinetic energy of the gas:

E_K = \frac{3}{2}kT

where

k = 1.38066 × 10−23 J/K is the Boltzmann's constant

Substituting,

E_K = \frac{3}{2}(1.38066\cdot 10^{-23} J/K)(239.2 K)=4.95\cdot 10^{-21} J

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A girl and a boy are riding on a merry go round that is turning at a constant rate. The girl is near the outer edge, and the boy
galina1969 [7]

Answer:

The girl has greater tangential acceleration

Explanation:

The angular acceleration (\alpha) of the merry go round is equal to the rate of the change of the angular velocity, \omega:

\alpha = \frac{d\omega}{dt}

Since all the points of the merry go round complete 1 circle in the same time, the angular velocity of each point of the merry go round is the same, and so all the points also have the same angular acceleration.

The tangential acceleration instead is given by

a_t = \alpha r

where

\alpha is the angular acceleration

r is the distance from the centre of the merry go round

Since the girl is near the outer edge and the boy is closer to the centre, the value of r for the girl is larger than for the boy, so the girl has greater tangential acceleration.

5 0
3 years ago
A long, East-West-oriented power cable carrying an
Alla [95]

Answer:

200A

Explanation:

Given that

the distance between earth surface and power cable d = 8m

when the current is flowing through cable , the magnitude flux density at the surface is 15μT

when the current flow throught is zero the magnitude flux density at the surface is 20μT

The change in flux density due to the current flowing in the power cable is

B = 20μT - 15μT

B =5μT -----(1)

The expression of magnitude flux density produced by the current carrying cable is

B=\frac{\mu_0I}{2\pi d}-----(2)

Substitute the value of flux density

B from eqn 1 and eqn 2

\frac{\mu_0I}{2\pi d}=5\times 10^-^6\\\\\frac{(4\pi \times 10^-^7)I}{2 \pi (8)} =5\times 10^-^6\\\\I=200A

Therefore, the magnitude of current I is 200A

8 0
2 years ago
Sports car goes from a velocity of zero to a velocity of 12 m/s East in two seconds. What is the cars acceleration?
bogdanovich [222]

Answer:

6 m/sec

Explanation:

12/2=6

7 0
3 years ago
Electromagnetic radiation of 8.12×10¹⁸ Hz frequency is applied on a metal surface and caused electron emission. Determine the wo
klemol [59]

Answer:

The work function ϕ of the metal = 53.4196 x 10⁻¹⁶ J      

Explanation:

When light is incident on a photoelectric material like metal, photoelectrons are emitted from the surface of the metal. This process is called photoelectric effect.

The relationship between the maximum kinetic energy (E_{k}) of the photoelectrons to the frequency of the absorbed photons (f) and the threshold frequency (f₀) of the photoemissive metal surface is:

                                        E_{k} = h(f − f₀)

                                        E_{k} = hf - hf₀

E is the energy of the absorbed photons:  E = hf

ϕ is the work function of the surface:  ϕ = hf₀

                                        E_{k} = E - ϕ

Frequency f = 8.12×10¹⁸ Hz

Maximum kinetic energy E_{k} = 4.16×10⁻¹⁷ J  

Speed of light  c = 3 x 10⁸ m/s

Planck's constant h = 6.63 × 10⁻³⁴ Js                                

                                        E = hf = 6.63 × 10⁻³⁴ x 8.12×10¹⁸

                                        E = 53.8356 x 10⁻¹⁶ J

from E_{k} = E - ϕ ;

                                        ϕ = E - E_{k}

                                        ϕ = 53.8356 x 10⁻¹⁶ - 4.16×10⁻¹⁷

                                        ϕ = 53.4196 x 10⁻¹⁶ J

The work function of the metal ϕ = 53.4196 x 10⁻¹⁶ J      

4 0
3 years ago
A projectile is launched from ground level to the top of a cliff which is 195 m away and 155 m high. If the projectile lands on
muminat

Answer:

66.02m/s

Explanation:

the equation describing the distance covered in the horizontal direction is

x=ucos\alpha t-(1/2)gt^{2} but the acceleration in the horizontal path is zero, hence we have

x=ucos\alpha t

Since the horizontal distance covered is 155m at 7.6secs, we have ucos\alpha =\frac{155}{7.6} \\ucos\alpha =20.38.............equation 1

Also from the vertical path, the distance covered is expressed as

y=usin\alpha t-(1/2)gt^{2}

since the horizontal distance covered in 7.6secs is 195m, then we have

y=usin\alpha t-(1/2)gt^{2}\\y=7.6usin\alpha -4.9(7.6)^{2}\\478.02=7.6usin\alpha \\usin\alpha =62.9...........equation 2

Hence if we divide both equation 1 and 2 we arrive at

\frac{usin\alpha }{ucos\alpha } =\frac{62.9}{20.38} \\tan\alpha =3.08\\\alpha =tan^{-1}(3.08)\\\alpha =72.02^{0}\\

Hence if we substitute the angle into the equation 1 we have

ucos72.02=20.38\\u=66.02m/s

Hence the initial velocity is 66.02m/s

3 0
3 years ago
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