Answer:
the weight of the ball is w = 51.94 N ( mass = 5.3 kg)
Explanation:
Following Newton's second law:
net force = mass * acceleration = weight/gravity * acceleration
then denoting 1 and 2 as the first and second lift
F₁ - w= w/g *a₁
F₂ -w = w/g *a₂ = w/g * 2.07a
dividing both equations
(F₂- w)/(F₁ -w)= 2.07
(F₂- w) = 2.07 * (F₁ -w)
1.07*w = 2.07*F₁ - F₂
w = (2.07*F₁ - F₂ )/ 1.07
replacing values
w = (2.07*61.1 N - 70.9 N )/ 1.07 = 51.94 N
then the weight of the ball is w = 51.94 N ( mass = 5.3 kg)
Answer:
x(t) = - 6 cos 2t
Explanation:
Force of spring = - kx
k= spring constant
x= distance traveled by compressing
But force = mass × acceleration
==> Force = m × d²x/dt²
===> md²x/dt² = -kx
==> md²x/dt² + kx=0 ------------------------(1)
Now Again, by Hook's law
Force = -kx
==> 960=-k × 400
==> -k =960 /4 =240 N/m
ignoring -ve sign k= 240 N/m
Put given data in eq (1)
We get
60d²x/dt² + 240x=0
==> d²x/dt² + 4x=0
General solution for this differential eq is;
x(t) = A cos 2t + B sin 2t ------------------------(2)
Now initially
position of mass spring
at time = 0 sec
x (0) = 0 m
initial velocity v= = dx/dt= 6m/s
from (2) we have;
dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)
put t =0 and dx/dt = v(0) = -6 we get;
-2A sin 2(0)+2Bcos(0) =-6
==> 2B = -6
B= -3
Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get
x(t) = - 6 cos 2t
==>
Answer:
A
Explanation:
the metal lunchbox has a higher conductivity
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I believe that the answer to this would be B
Hope this helped
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.