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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%

Question

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Respuesta :

SaniShahbaz SaniShahbaz · Answer 1 · 2019-09-29T17:12:34+02:00

Answer:

1) Joule

2) kgm²/s²

3) 10%

Explanation:

Question 1)

Luminosity is defined as the total amount of energy released from an object in a unit amount of time.

This means Luminosity is the ratio of energy and time.

Unit of Energy is Joules and Unit of Time is seconds, so unit of Luminosity would be Joule per second or Watts.

So, from the given options, the unit which would be used in Luminosity calculations is Joule.

Question 2)

Work = Force x Distance

SI unit of Work is Joule

SI unit of Force is Newton or kgm/s²

SI unit of Distance is meter(m)

Unit of Work = Unit of Force x Unit of Distance

Joule = kgm/s² x m

Joule = kgm²/s²

This means, the correct description of joule in terms of SI base units is kgm²/s²

Question 3)

Value of gravitational constant from the experiment = [tex]7.34 \times 10^{-11}[/tex] Nm²/kg²

The accepted or actual value of the gravitational constant = [tex]6.67 \times 10^{-11}[/tex] Nm²/kg²

We need to calculate the percentage error. Percentage error is defined as:

[tex]\text{Percentage Error}=\frac{\text{New Value - Original Value}}{\text{Original Value}} \times 100\%[/tex]

Here, original value is the accepted value. Using the values in this formula, we get:

[tex]\frac{7.34 \times 10^{-11}-6.67 \times 10^{-11}}{6.67 \times 10^{-11}}\\= 10\%[/tex]

Thus, this means the percentage error in the calculation of gravitational constant was 10%.