Which of the following would be used in luminosity calculations?

candela

joule

kilogram

ampere

2. Work, measured in joules, is force × distance.

If the unit of force is kg⋅m/s2 , then which of the following is the correct description of joule in terms of SI base units?

kg⋅m/s3

kg⋅m/s2

kg⋅m/s

kg⋅m2/s2

3.In an experiment, you calculate Newton's universal gravitational constant to be 7.34×10−11 N⋅m2/kg2 . The accepted value is 6.67×10−11 N⋅m2/kg2 .

What is your percent error?

0.01%

0.1%

5%

10%

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 ([tex]\frac{J}{s}[/tex]), which is equal to Watts ([tex]W[/tex]).

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 [tex]W[/tex] is expressed as force [tex]F[/tex] multiplied by the distane [tex]d[/tex] :

[tex]W=F.d[/tex]

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

If we input the units we will have:

[tex]W=(kgm/s^{2})(m)[/tex]

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

3) The formula to calculate the percent error is:

[tex]\% error=\frac{|V_{exp}-V_{acc}|}{V_{acc}} 100\%[/tex]

Where:

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

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

[tex]\% 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\%[/tex]

[tex]\% error=10.04\% \approx 10\%[/tex] This is the percent error