Consider the expression [tex]2 \pi \sqrt{\frac{l}{g}}[/tex].
where ℓ is length and g is gravitational acceleration in units of length divided by time squared.
Evaluate its units.
1. s²
2. [tex](\frac{s}{m})^2[/tex]
3. [tex](\frac{m}{s})^2[/tex]
4. [tex]\frac{m}{s}[/tex]
5. s
6. m²
7. m
8. [tex]\frac{s}{m}[/tex]

Respuesta :

Answer:

Step-by-step explanation:

lets start from the expression [tex]2\pi \sqrt{\frac{l}{g} }[/tex], first remember that [tex]2\pi[/tex] is a constant with no units, so we do not need to consider it for this problem, so lets just think about [tex]\frac{l}{g}[/tex]

Next lets replace the units, l is the length which is represented in metters by international units system

g is the gravitational constants, so the units are lengt divided by time squared, since time is in seconds we have [tex]\frac{m}{s^{2} }[/tex]

so pluging that into our formula we have [tex]\sqrt{\frac{m}{\frac{m}{s^{2} } } }[/tex]

Now see we are dividing by a fraction, remember that to divide by a fraction is the same as to multiply by the reciprocal fraction, so we just flip the fraction and we get [tex]\sqrt{m*\frac{s^{2} }{m} }[/tex]

then if we perform the multiplication we have inside the radical we get [tex]\sqrt{\frac{m*s^{2} }{m} }[/tex]

Then since we are multiplying and dividing by m, the m's just cancel and we have [tex]\sqrt{s^{2} }[/tex]

Now remember that square and square root are inverse operations, so they cancel each other and we are left with just [tex]s[/tex], which means seconds, so the answer would be option 5

hope that helps, good bye :)