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does anyone know how to do these?1.the mass of beryllium nucleus (4,7Be) is 1.1652 x...

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wong23456 | Student, Grade 11 | eNoter

Posted August 1, 2012 at 7:15 AM via web

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does anyone know how to do these?

1.the mass of beryllium nucleus (4,7Be) is 1.1652 x 10^-26 kg. What is the mass defect of the beryllium nucleus? What is the binding energy of the berylliuy nucleus? What is the binding energy per nucleon on the beryllium nucleus. 

2.What is the mass of a proton travelling atv = 0.80c?

3.At what speed v will the mass of an object be 10% greater than its rest mass?

4.You have just won a trip to travel to a distance star. You will be travelling at aspeed of 2.60 X 10^8 m/s and the star is 100 light-years away. What would you measure this distance to be?

5.Now your friend is travelling at a velocity of 0.80 c towards you. You arestationary on earth. Your friend throws a baseball forward at a velocity of 0.60 c(relative to your friend). What is the velocity of the baseball with respect to you?



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eriqah | Student , Grade 10 | eNoter

Posted August 5, 2012 at 11:04 AM (Answer #1)

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are u dat smart by asking these q'stns ... try answering then...

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted October 10, 2013 at 7:03 AM (Answer #2)

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(4,7)Be has 4 protons and 7-4 =3 neutrons. The mass defect of Be is the difference between the total mass of its nucleons taken separately and the mass of Be nucleus.

`Delta(m) = 4*m_p +3*m_n -m(Be) =(4*1.6726 +3*1.6749 -11.652)*10^-27 =0.0631*10^-27 kg =`

`=(0.0631*10^-27)/(1.66*10^-27) =0.038 "amu"`


When travelling at very high speed the mass becomes greater with respect to the rest mass

`m =m_0/sqrt(1-v^2/c^2) =(1.6726*10^-27)/sqrt(1-0.8^2) =2.7877*10^-27 kg= 1.679 amu` 3)

As described above the mass becomes grater with increasing speed.

`m/m_0 =1/sqrt(1-v^2/c^2)`

`m =m_0 +0.1*m_0=1.1*m_0`

`1.1^2 =1/(1-v^2/c^2)`

`1-v^2/c^2 =0.8264`

`v^2/c^2 =0.1735`

`v =0.416*c =0.416*3*10^8 =1.25*10^8 m/s`


In the moving reference frame the lengths contract.

`L =L_0*sqrt(1-v^2/c^2) =100*sqrt(1-(2.6/3)^2) =49.99 "lightyears"`


The composition of speeds in relativity is done as

`s =(v+u)/(1+v*u/c^2) =(0.8+0.6)/(1+(0.8*0.6)/1) =0.946*c`

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