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Physics: Post your doubts here!

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Re: Physics questions? please?

Question 10 Please.

Thanks!


u can easily learn the formulas and definations being asked in this question from the application support booklet on page 22! its in detail and easy chek it.
 
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first of all regarding no. 11 f=ma the weight of the substance is 20kg an its mass is 2 as g=10 in this case
two forces in opposite direction so 10-4=6
6=ma
6=2a
a=3
regarding 13
first thing what could be the minimum value of cos its o when all the forcse all prependicular to each other so by resolution when all forcse will get 0 as cos will get 0 so A is the answer
if you still have a confusion do tell me
 
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another question when we make a solution of two miscible liquids the boling point of the solution is effected...now ignore that whether it increases or decreases but the whole sol. has a uniform boiling point...if it is so then how we will perform the process of fractional distillation????
Chemistry question, i believe this is a application question, it's about successive extraction, pg 82. In chem app booklet.
They might have same bp, but not solubility in certain solvent.
 
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a beam of monochromatic lite of wavelengh 630nm trnsports energy at the rate of 0.25mW. calculate the no. of photons passing a given cross section of beam each second.(planck cnstant=6.6*10-34,speed of light=3.0*108)

ans is [8.0*1014] ..................*10 stands for exp
 
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Chemistry question, i believe this is a application question, it's about successive extraction, pg 82. In chem app booklet.
They might have same bp, but not solubility in certain solvent.
i did o lvls and thn doing fsc and dnt have tht book...cn u pls upload tht page...:)
 
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a beam of monochromatic lite of wavelengh 630nm trnsports energy at the rate of 0.25mW. calculate the no. of photons passing a given cross section of beam each second.(planck cnstant=6.6*10-34,speed of light=3.0*108)

ans is [8.0*1014] ..................*10 stands for exp
Use E=(hc)/lambda. where E=(6.63*10^-34*3*10^8)/(630*10^-9) E=3.157*10^-19 J is the energy of one photon.

You have the beam transporting energy at the rate of 0.25mW=0.25mJ So Divide 0.25*10^-3 by 3.157*10^-10 to obtain the number of photons. You get 7.92*10^14 photons being transported.
 
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