If v is a velocity, l a length, and v a fluid property having dimensions of l2t-1, which of the following combinations are qualitatively dimensionless (show work for a through d): (a) v l v(b) vl/v(c) v2v(d) v / (l v)

(b) Vl/v

Explanation:

The dimension for V is LT-1

The dimension for Length is L

The dimension for time is T

The fluid property v has a dimension of L2T-1, which is equal to square of the length divided by time.

for (a) Vlv

LT-1×L× L2T-1

= L^4T^-2

(b) Vl/v

LT^-1 ×L/ L^2 T^-1

=1. ; this is dimensionless.

(c)V2v

(LT^-1)^2 × L^2 T^-1

= L^4 T^-3

(d) V/(lv)

LT^-1/(LL^2 T^-1)

= L^-2

i don’t know what the answer is i wish i

the motion of a pendulum is a classic example of mechanical energy conservation. a pendulum consists of a mass (known as a bob ) attached by a string to a pivot point. as the pendulum moves it sweeps out a circular arc, moving back and forth in a periodic fashion.