A box with a mass of 0.6-kg falls from a very tall building. While free-falling it is subjected to a frictional drag force given by Fdrag = bv2, where v is the speed of the object and b=5.5 N•s2/m2 What is the terminal speed that the box will reach?

the terminal speed that the box will reach v = 1.0355 m/s

Explanation:

The drag force is the resultant force component that acts in the relative motion direction of the body.

Here we have Fdrag = bv²

But Fdrag = mg

Therefore F = 0.6 × 9.81 =5.886 N

v² = Fdrag / b = 5.886 N/(5.5 N s²/m²)

v = 1.0355 m/s

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary