Question

The kinetic energy (KE) of a motorcycle of mass (m) 95 kg (correct to 10 kgs) traveling at an average velocity (v) of 20 ms-1 (correct to 1 ms-1) is given by

KE = 1 / 2 mv2.

Calculate the range of the upper and lower bounds of the kinetic energy of the motorcycle.

________ joules ≤ KE < ________joules.

Answer 1

Answer:

Simplify the given expression. (5+√7) (2+√5)