Question

A small, spherical bead of mass 2.70 g is released from rest at t = 0 from a point under the surface of a viscous liquid. The terminal speed is observed to be vT = 1.82 cm/s. (a) Find the value of the constant b in the equation = −b.

Answer 1

Answer:

The value of constant b will be 1.455  

Explanation:

We have given mass  of the bead $m=2.70gram=2.70\times 10^{-3}kg$

Acceleration due to gravity $g=9.81m/sec^2$

Terminal velocity $v_t=1.82cm/sec=1.82\times 10^{-2}m/sec$

We have to find the value of constant b

We know that constant b is given by

$b=\frac{mg}{v_t}=\frac{2.70\times 10^{-3}\times 9.81}{1.82\times 10^{-2}}=1.455$

So the value of constant b will be 1.455