A 3 mm inside diameter tube is placed in a fluid with a surface tension of 600 mN/m and density of 3.7 g/cm3. The contact angle
1 answer:
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
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Answer:
15.64 KN
Explanation:
mass of the elevator cab with a single occupant= 2300 kg
acceleration relative to the cab = 6.80 m/s^2
acceleration of the coin relative to the cab in the opposite direction of motion of cab so we can consider it as a= -6.80 m/s^2
The acceleration of elevator cab relative to the ground
now we can say that
=-6.80+ a_{eg}= -9.8
[tex]a_{eg}=-9.8+6.80=-3.8
The forces that act on elevator cab are tension and gravitational, applying newtons second law
T- mg= ma_{eg}
Then the tension in the cable is
T= 2300(-3.8)+2300×9.8= 15640 N= 15.64 KN
therefore tension in the string will be 15.64 KN