Question

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%

Answer 1

1) joule

2)
`kgm^(2)/s^(2)`

3)
`10\%`

Step-by-step 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 (
`(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=(|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=(|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