Lf a dropped hammer's velocity increases from 0.Omis to 15.0 m/s in 4.04s
1 answer:
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Answer:
a)-1.014x J
b)3.296 x J
Explanation:
For Sphere A:
mass 'Ma'= 47kg
xa= 0
For sphere B:
mass 'Mb'= 110kg
xb=3.4m
a)the gravitational potential energy is given by
= -GMaMb/ d
= - 6.67 x
x 47 x 110/ 3.4 => -1.014x
J
b) at d= 0.8m (3.4-2.6) and =-1.014x
J
The sum of potential and kinetic energies must be conserved as the energy is conserved.
+
=
+
As sphere starts from rest and sphere A is fixed at its place, therefore is zero
=
+
The final potential energy is
= - GMaMb/d
Solving for ' '
=
+ GMaMb/d => -1.014x
+ 6.67 x
x 47 x 110/ 0.8
= 3.296 x
J
Answer:
As the concentration of a solute in a solution increases, the freezing point of the solution <u><em>decrease </em></u>and the vapor pressure of the solution <em><u>decrease </u></em>.
Explanation:
Depression in freezing point :
where,
=depression in freezing point =
= freezing point constant
m = molality ( moles per kg of solvent) of the solution
As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:
- If molality of the solution in high the depression in freezing point of the solution will be more.
- If molality of the solution in low the depression in freezing point of teh solution will be lower .
Relative lowering in vapor pressure of the solution is given by :
= Vapor pressure of pure solvent
= Vapor pressure of solution
= Mole fraction of solute
Vapor pressure of the solution is inversely proportional to the mole fraction of solute.
- Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
- lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.