$\therefore$ Volume of 1 mole of hydrogen atoms = 6.023 × 1023 × 0.524 × 10-30
= 3.16 × 10-7m350 questions · timed · auto-graded
$\therefore$ Volume of 1 mole of hydrogen atoms = 6.023 × 1023 × 0.524 × 10-30
= 3.16 × 10-7m3$= 15072 = 1.5\times104\text{mm}^2$
$\Big(\frac{1}{100}\Big)\Big(\frac{1}{100}\Big)\Big(\frac{1}{100}\Big)\text{m}^3=\frac{1}{6^6}\text{m}^3=10^{-6}\text{m}^3$
6.032 N m-2
The given quantity is 0.0006032m2.
If the number is less than one, then the zeroes on the right of the decimal point (but left to the first non-zero) are insignificant. Hence, all three zeroes appearing before 6 are not significant figures. All zeros between two non-zero digits are always significant. Hence, the remaining four digits are significant figures.1N = 1kg m s-2
1kg = 103g 1m = 100cm = 102cm Putting above values, we get 6.67 × 10–11 N m2kg–2 = 6.67 × 10–11 × (1kg m s–2) (1m2) (1Kg–2) Solve and cancel out the units we get ⇒ 6.67 × 10–11 × (1kg–1 × 1m3 × 1s–2) Putting above values to convert Kg to g and m to cm ⇒ 6.67 × 10–11 × (103g)-1 × (102cm)3 × (1s–2) ⇒ 6.67 × 10–11 × 10-3g-1 × 106cm3 × (1s–2) ⇒ 6.67 × 10–8cm3 s–2 g–1 G = 6.67 × 10–11 N m2 (kg)–2 = 6.67 × 10–8 (cm)3s–2 g–1.0.2370g cm-3
Explanation:
The given quantity is 0.2370g cm-3. For a number with decimals, the trailing zeroes are significant. Hence, besides digits 2, 3 and 7, 0 that appears after the decimal point is also a significant figure.Radius
$31.0\text{cm},\theta=\frac{\pi}{6}$From length of the arc of a circle
$\text{l}=\text{r}\theta$$=31.0\times\frac{\pi}{6}$
$=16.2\text{cm}$
$[\text{M}^{\frac{3}{2}}\text{L}^{-1}\text{T}^{-2}]$
$[\text{M}^{\frac{1}{2}}\text{L}^{\frac{-3}{2}}\text{T}^{0}]$
Explanation:
$\text{p} = \text{q} \times 2\text{a} = (\text{AT}) \times \text{L} = [\text{M}^{0}\text{L}^1\text{T}^1\text{A}^1] $
Dimensions of elecric field
$\text{e}=\frac{\text{F}}{\text{q}}=\frac{\text{MLT}^{-2}}{\text{AT}}=[\text{MLT}^{-3}\text{A}^{-1}]$
$\phi=\vec{\text{E}}.\vec{\text{A}}=\text{EA}\cos\theta$
$\frac{\text{F}}{\text{q}}\text{A}\cos\theta=\Big[\frac{\text{MLT}^{-2}}{\text{AT}}\times\text{L}^2\times1\Big]$
$=[\text{M}^{1}\text{L}^3\text{T}^{-3}\text{A}^{-1}]$
Explanation:
Reynolds number and coefficient of friction, both are dimensionless.
$\text{L}=\frac{\text{Q}}{\text{m}}=\frac{\text{ML}^2\text{T}^{-2}}{\text{M}}=[\text{L}^{2}\text{T}^{-2}]$
Gravitational Potential $=\frac{\text{W}}{\text{m}}=\frac{\text{ML}^{2}\text{T}^{-2}}{\text{M}}=[\text{L}^2\text{T}^{-2}]$
1 curie $=3.7\times10^{10}$ disintegrations/ sec =
$\text{T}^{-1}$= Frequency $=\text{T}^{-1}$
Explanation:
$=\frac{\text{Momentum}}{\text{Area}\times\text{Time}}\frac{\text{MLT}^{-1}}{\text{L}^{2}\text{T}}=[\text{M}^{1}\text{L}^{-2}\text{T}^{-2}]$