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Greeley [361]
3 years ago
9

What units do chemists normally use for density of liquids and solids? for gas density? explain the differences?

Physics
1 answer:
anygoal [31]3 years ago
5 0

We can define density as per unit volume.

The SI unit to measure the density of solid, liquid and gas is kilogram per cubic metre (kg/m3) and in the centimetre–gram–second system of units (cgs unit) is gram per cubic centimetre (g/cm3). Gas density is very dependent of pressure and temperature whereas the density of solids and liquids is not so dependent, that is why the gas density is given at a standard temperature and pressure.

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particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine
AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is 1062.7\frac{m}{s^{2}}

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

F=ma (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}

with q1 and q2 the charge of the particles, r the distance between them and k the constant k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}. So:

F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}

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Using that value on (1) and solving for a

a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}

4 0
3 years ago
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

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Nesterboy [21]

Answer:

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Explanation:

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