Answer:
v = 1.49 m/s
Explanation:
given,
side of block = 200 mm = 0.2 m
Terminal speed of the block
[tex]v = \dfrac{hWsin\theta}{\mu\ A}[/tex]
W is the weight of the block = 1 kN = 1000 N
h is the thickness of the = 0.005 mm
A is the area of the block
θ is the angle of inclination, = 30°
A = 6 a²
A = 6 x 0.2² = 0.24 m²
Viscosity of the oil,μ = 7 x 10⁻³N.s/m²
now, inserting all the values
[tex]v = \dfrac{0.005\times 10^{-3}\times 1000\times \sin 30^{\circ}}{7\times 10^{-3}\times 0.24}[/tex]
v = 1.49 m/s
Hence, the terminal velocity is equal to v = 1.49 m/s
