Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

From the question we are told
If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,
Generally the equation Time spent is mathematically given as
T=\frac{d}{v}
Therefore

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

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In the part of the spectrum our eyes can detect (a spectrum is an arry of entities, as light waves or particles, ordered in accordance with the magnitudes of a common physical property, as wavelength or mass) Hope this helps you :D
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Answer:
Explanation:
1 )
We shall apply conservation of momentum law to solve the problem.
mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .
.5 x .2 = (1.5 + .5)V
V = .05 m /s
2 ) We shall use formula for velocity of object after elastic collision as follows
v₁ = 
m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.
Putting the values
= 
= - .66 m /s
Since the sign is negative so it will be in opposite direction .