Answer: The height of the fluid rise is 0.01m
Explanation:
Using the equation
h = (2TcosѲ )/rpg
h= height of the fluid rise
diameter of the tube =3mm
radius of the tube= 3/2 =1.5mm=0.0015
T= surface tension = 600mN/m=0.6N/m
Ѳ = contact angle = 
C
p= density =3.7g/cm3= 3700kg/m3
g= acceleration due to gravity =9.8m/s2
h = ( 2*0.6*0.5)/(0.0015*3700*9.8)
h = 0.6/54.39
h= 0.01m
Therefore,the height of the fluid rise is 0.01m
I'll tell you how I look at this, although I may be missing something important.
Position = x(t) = 0.5 sin(pt + p/3)
Speed = position' = x'(t) = 0.5 p cos(pt + p/3)
Acceleration = speed' = position ' ' = x ' '(t) = -0.5 p² sin(pt + p/3)
At (t = 1.0),
x ' '(t) = -0.5 p² sin( 4/3 p )
In order to evaluate this, don't I still have to know what 'p' is ? ?
I don't think it can be evaluated with the information given in the question.
Answer:
Explanation:
We shall apply law of conservation of momentum in space to know the velocity of combination after the impact
m₁v₁ = m₂v₂
.1 x 4 = ( 1 + .1 ) v₂
v₂ = .3636 m /s
1 )
Kinetic energy of the combination
= 1/2 x 1.1 x ( .3636)²
= 7.3 x 10⁻² J
2 )
Initial kinetic energy of the system
= 1/2 x 0.1 x 4²
= 0.8 J
Final kinetic energy of the system = 7.3 x 10⁻²
Loss of energy = .8 - .073
= .727 J
This energy was converted into internal energy of the system .
3 )
increase in entropy = dQ / T
Here dQ = .727 J
T = 300 ( Constant )
dQ / T = 2.42 X 10⁻³ J/K
Answer:
Probs paper cuz the rest are metals
