Question

Which of the following would be used in luminosity calculations?

Answer 1

Answer:

1) joule

2) $kgm^{2}/s^{2}$

3) $10\%$

Explanation:

1) Luminosity is the amount of light emitted (measured in Joule) by an object in a unit of time (measured in seconds). Hence in SI units luminosity is expressed as joules per second ($\frac{J}{s}$), which is equal to Watts ($W$).

This amount of light emitted is also called radiated electromagnetic power, and when this is measured in relation with time, the result is also called radiant power emitted by a light-emitting object.

Therefore, if we want to calculate luminosity the Joule as a unit will be used.

2) Work $W$ is expressed as force  $F$ multiplied by the distane  $d$ :

$W=F.d$

Where force has units of  $kgm/s^{2}$ and distance units of $m$.

If we input the units we will have:

$W=(kgm/s^{2})(m)$

$W=kgm^{2}/s^{2}$  This is 1Joule ($1 J$) in the SI system, which is also equal to $1 Nm$

3) The formula to calculate the percent error is:

$\% error=\frac{|V_{exp}-V_{acc}|}{V_{acc}} 100\%$

Where:

$V_{exp}=7.34 (10)^{-11} Nm^{2}/kg^{2}$ is the experimental value

$V_{acc}=6.67 (10)^{-11} Nm^{2}/kg^{2}$ is the accepted value

$\% error=\frac{|7.34 (10)^{-11} Nm^{2}/kg^{2}-6.67 (10)^{-11} Nm^{2}/kg^{2}|}{6.67 (10)^{-11} Nm^{2}/kg^{2}} 100\%$

$\% error=10.04\% \approx 10\%$ This is the percent error