Turn on thread page Beta
    • Thread Starter
    Offline

    10
    ReputationRep:
    The present mass of the Sun is 2.0 × 10^30 kg. The Sun emits radiation at an average rate of 3.8 × 10^26 J s–1. Calculate the time in years for the mass of the Sun to decrease by one millionth of its present mass.

    From markscheme: "E = mc2
    energy = 2.0*10^30 *10^-6 *(3.0108 )2"

    why 10^-6? How is that a change in mass? because the equation the markscheme uses is change in energy = change in mass * speed ^2
    Offline

    11
    ReputationRep:
    (Original post by MrToodles4)
    The present mass of the Sun is 2.0 × 10^30 kg. The Sun emits radiation at an average rate of 3.8 × 10^26 J s–1. Calculate the time in years for the mass of the Sun to decrease by one millionth of its present mass.

    From markscheme: "E = mc2
    energy = 2.0*10^30 *10^-6 *(3.0108 )2"

    why 10^-6?

    EDIT: wait nvm, its for a millionth of the mass isn't it..
    You have to find the time and not the energy.
    Offline

    11
    ReputationRep:
    (Original post by MrToodles4)
    The present mass of the Sun is 2.0 × 10^30 kg. The Sun emits radiation at an average rate of 3.8 × 10^26 J s–1. Calculate the time in years for the mass of the Sun to decrease by one millionth of its present mass.

    From markscheme: "E = mc2
    energy = 2.0*10^30 *10^-6 *(3.0108 )2"

    why 10^-6? How is that a change in mass? because the equation the markscheme uses is change in energy = change in mass * speed ^2
    2.0*10^30 *10^ is one million of the mass.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by WiSi)
    You have to find the time and not the energy.
    Yeah but I thought delta m is the change in mass. I can find the time afterwards by doing p=e/t so e/p = t
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by WiSi)
    2.0*10^30 *10^ is one million of the mass.
    Is that considered 'change in mass then'? i thought it would be the mass to decrease to get to one millionth of the mass
    Offline

    11
    ReputationRep:
    (Original post by MrToodles4)
    Yeah but I thought delta m is the change in mass. I can find the time afterwards by doing p=e/t so e/p = t
    What's your doubt about the problem ?
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by WiSi)
    What's your doubt about the problem ?
    Basically normally change in mass would be final mass - initial mass right? So in this case it it right to assume that the DECREASE of one millionth IS the change in mass?
    Offline

    11
    ReputationRep:
    (Original post by MrToodles4)
    Basically normally change in mass would be final mass - initial mass right? So in this case it it right to assume that the DECREASE of one millionth IS the change in mass?
    It should be its actual mass minus one million of his mass.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by WiSi)
    It should be its actual mass minus one million of his mass.
    The markscheme does not do that though...

    3b: http://pmt.physicsandmathstutor.com/...y%204%20MS.pdf

    Question 3B: http://pmt.physicsandmathstutor.com/...y%204%20QP.pdf
    Offline

    9
    ReputationRep:
    (Original post by MrToodles4)
    The markscheme does not do that though...
    The question states that the mass decreases BY one millionth of its mass.

    so delta m = 1x10^-6 times the mass

    It does not decrease TO one millionth of the mass
    Offline

    11
    ReputationRep:
    I understand the solution.
    So, the Sun have a mass of 2.0 × 10^30 kg.
    Your problem ask time sun needs to decrease by one millionth of its present mass.
    So you are going to calculate the Energy the Sun lose, while losing one millionth of its present mass.
    For this reason you have to use the value of the mass the sun lose, and not the mass that remains in the Sun.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: February 27, 2018

University open days

  • University of Chichester
    Multiple Departments Undergraduate
    Thu, 25 Oct '18
  • Norwich University of the Arts
    Undergraduate Open Days Undergraduate
    Fri, 26 Oct '18
  • University of Lincoln
    Open Day Undergraduate
    Sat, 27 Oct '18
Poll
Who do you think it's more helpful to talk about mental health with?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.