Answer:
A) 0.957 J
Explanation:detailed explanation and calculation is shown in the image below
Shear stress = 1.0 N/m² (Pa)
For water, the dynamic viscosity = 10⁻³ Pa-s at 20°C.
The velocity gradient required = (Shear stress)/(Dynamic viscosity)
= (1.0 Pa)/( 10⁻³ Pa-s)
= 10³ 1/s
Answer: 10³ s⁻¹
Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
Answer:2.53*10^-10F
Explanation:
C=£o£r*A/d
Where £ is the permitivity of a constant
£o= 8.85*10^-12f/m
£r=6.3
A=150mm^2=0.015m^2
d=3.3mm= 0.0033m
C=8.85*10^-12*6.3*0.015/0.0033
C=8.85*6.3*10^-12*0.015/0.0033
C=55.755*0.015^-12/0.003
C=8.36/3.3*10^-13+3
C=2.53*10^-10F
