3. 500ml of 60oC water at is added to a beaker. Water’s specific heat is 4.186 J/g °C. You have 40ml of and unknown substance th
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Answer:
Mass of the climber = 69.38 kg
Explanation:
Change in length
Load, P = m x 9.81 = 9.81m
Young's modulus, Y = 0.37 x 10¹⁰ N/m²
Area
Length, L = 15 m
ΔL = 5.1 cm = 0.051 m
Substituting
Mass of the climber = 69.38 kg
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