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3 years ago

Use the dimensional analysis and check the correctness of given equation:- PV= nRT

Physics

1 answer:

Harman [31]3 years ago

6 0

PV=nRT

Here

P=Pressure

V=Volume

n=Molarity

R=universal gas constant

T=Temperature.

LHS

$\\ \tt\bull\leadsto PV$

$\\ \tt\bull\leadsto [ML^2T^{-2}][M^0L^3T^0]$

$\\ \tt\bull\leadsto [ML^5T^{-2}]$

RHS

$\\ \tt\bull\leadsto nRT$

$\\ \tt\bull\leadsto [M^0L^{3}T^0][M^1 L^2 T^{-2}K^{-1}][M^0L^0T^0K^1]$

$\\ \tt\bull\leadsto [ML^5T^{-2}]$

LHS=RHS

hence verified

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Halleys comet has period of 75.3 years. Using Kepler’s third law, find it’s semimajor axis expressed in astronomical units?

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Answer: 17.83 AU

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. 

$T^{2}\propto a^{3}$ (1)

Talking in general, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite, comet) orbiting a greater body in space with the size $a$ of its orbit.

However, if $T$ is measured in years, and $a$ is measured in astronomical units (equivalent to the distance between the Sun and the Earth: $1AU=1.5(10)^{8}km$ ), equation (1) becomes:

$T^{2}=a^{3}$ (2)

This means that now both sides of the equation are equal.

Knowing $T=75.3years$ and isolating $a$ from (2):

$a=\sqrt[3]{T^{2}}=T^{2/3}$ (3)

$a=(75.3years)^{2/3}$ (4)

Finally:

$a=17.83AU$ (5)

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3 years ago

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3 years ago

The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC

enyata [817]

Answer:

The y-component of the electric force on this charge is $F_y = -1.144\times 10^{-6}\ N.$

Explanation:

Given:

Electric field in the region, $\vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.$
Charge placed into the region, $q = 10.4\ nC = 10.4\times 10^{-9}\ C.$

where, $\hat i,\ \hat j$ are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

$\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.$

Thus, the y-component of the electric force on this charge is $F_y = -1.144\times 10^{-6}\ N.$

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