Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 1 - Sections 1.1-1.8 - Exercises - Cumulative Problems: 109


$N=\frac{kg\times m}{s^{2}}$ kN and pN are the most suitable.

Work Step by Step

1. $F= m\times a= \frac{m\times V}{t}=\frac{m\times l}{t^{2}}$ SI base units that force consists of are length -(m); time -(s); mass - (kg) 2. We convert speed from mph to m/s using conversion: $1 mile = 1.6093km$ $1km=1000m$ $1h=3600s$, then $85.5115 km/h = 88511.5 m/h = 24.5865 m/s$ 3. Calculate force: $F = \frac{10,000kg \times 24.5865 m/s}{1s}=245,865 N =245.865$ $ kN$ 4. Calculate next force without conversion: $F = \frac{10^{-20} kg \times 3 \times 10^{8} m/s}{1s}=3 \times 10^{-12} N =3$ $ pN$
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