Select all the correct answers.

Two cylinders A and B at the same temperature contain the same quantity of the same kind of gas. Cylinder A has three times the

astraxan [27]

Answer:

so pressure in A must be one third the pressure in B

Explanation:

We shall apply gas law to the cylinders A and B . Since their quantity are same so their no of mole will also be same .

For cylinder A

Temperature T , volume 3V , pressure P₁ , no of mole = n

so

P₁ X 3V = n R T

For cylinder B

Temperature T , volume V , pressure P₂ ,no of mole = n

so

P₂ X V = n R T

From the two equation above

P₁ X 3V = P₂ X V

$\frac{P_1}{P_2}=\frac{1}{3}$

P₁ = P₂ / 3

so pressure in A must be one third the pressure in B

3 0

3 years ago

Fill in the blanks ....... combine features of both web apps and native apps

natta225 [31]

Answer:

yvutvuvutvtyvyvtvytvtyvtvytvytvy

3 0

3 years ago

1.Do parallel universe exist? How?

Harlamova29_29 [7]

Answer:

Some physicists believe in a flatter version of multiple universes. ... But if the universe began at a finite point, as nearly every physicist agrees that it did, an alternate version of you likely doesn't exist, according to astrophysicist Ethan Siegel's 2015 Medium article.

7 0

2 years ago

A satellite moves in a circular orbit at a constant speed v0 around Earth at a distance R from its center. The force exerted on

postnew [5]

Answer:

vT = v0/3

Explanation:

The gravitational force on the satellite with speed v0 at distance R is F = GMm/R². This is also equal to the centripetal force on the satellite F' = m(v0)²/R

Since F = F0 = F'

GMm/R² = m(v0)²/R

GM = (v0)²R (1)

Also, he gravitational force on the satellite with speed vT at distance 3R is F1 = GMm/(3R)² = GMm/27R². This is also equal to the centripetal force on the satellite at 3R is F" = m(vT)²/3R

Since F1 = F'

GMm/27R² = m(vT)²/3R

GM = 27(vT)²R/3

GM = 9(vT)²R (2)

Equating (1) and (2),

(v0)²R = 9(vT)²R

dividing through by R, we have

9(vT)² = (v0)²

dividing through by 9, we have

(vT)² = (v0)²/9

taking square-root of both sides,

vT = v0/3

6 0

3 years ago

A plane monchromatic radio wave (lambda = 0.3 m) travels in vacuum along the positive x-axis, with an intensity I = 45 W/m2. Sup

Harrizon [31]

Answer:

$-5.78414\times 10^{-6}\ T$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

$\lambda$ = Wavelength = 0.3 m

T = Time period

f = Frequency

$E_0$ = Electric field

Intensity of electric field is given by

$I=\dfrac{1}{2}c\epsilon_0E_0^2\\\Rightarrow E_0=\sqrt{\dfrac{2I}{c\epsilon_0}}\\\Rightarrow E_0=\sqrt{\dfrac{2\times 45}{3\times 10^{8}\times 8.85\times 10^{-12}}}\\\Rightarrow E_0=184.11492\ N/C$

Magnetic field is given by

$B_0=\dfrac{E_0}{c}\\\Rightarrow B_0=\dfrac{184.11492}{3\times 10^8}\\\Rightarrow B_0=6.13716\times 10^{-7}\ T$

$k=\dfrac{2\pi}{\lambda}\\\Rightarrow k=\dfrac{2\pi}{0.3}\\\Rightarrow k=20.94\ /m$

$f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^{8}}{0.3}\\\Rightarrow f=1\times 10^{9}\ Hz$

$T=\dfrac{1}{f}\\\Rightarrow T=10^{-9}$

$\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{10^{-9}}\\\Rightarrow \omega=6283185307.17958\ rad/s$

Magnetic field in the z direction is given by (x=0)

$B_z=B_0(kx-\omega t)\\\Rightarrow B_z=6.13716\times 10^{-7}\times (0-6283185307.17958\times 1.5\times 10^{-9})\\\Rightarrow B_z=-5.78414\times 10^{-6}\ T$

The magnetic field at the origin is $-5.78414\times 10^{-6}\ T$

4 0

3 years ago