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

The value of constant b will be 1.455

Step-by-step explanation:

We have given mass of the bead
`m=2.70gram=2.70* 10^(-3)kg`

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

Terminal velocity
`v_t=1.82cm/sec=1.82* 10^(-2)m/sec`

We have to find the value of constant b

We know that constant b is given by

`b=(mg)/(v_t)=(2.70* 10^(-3)* 9.81)/(1.82* 10^(-2))=1.455`

So the value of constant b will be 1.455