The angular momentum is defined as,

Acording to this text we know for conservation of angular momentum that

Where
is initial momentum
is the final momentum
How there is a difference between the stick mass and the bug mass, we define that
Mass of the bug= m
Mass of the stick=10m
At the point 0 we have that,

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity
vector from the point of reference (O), and ve is the velocity
At the end with the collition we have

Substituting




Applying conservative energy equation we have


Replacing the values and solving

Substituting
l=\frac{13}{0.54(9.8)}

Answer:1 because
Explanation: it’s pointing to the earth and gravity
Pulls things down to earth
Answer:
depth of well is 163.30 m
Explanation:
Given data
speed of sound = 343 m/s
timer = 6.25 s
to find out
depth of well
solution
let us consider depth d
so equation will be
depth = 1/2 ×g ×t² ..............1
and
depth = velocity of sound × time .................2
here we have given time 6.25 that is sum of 2 time
when stone reach at bottom that time
another is sound reach us after stone strike on bottom
so time 1 + time 2 = 6.25 s
so from equation 1 and 2 we get
1/2 ×g ×t² = velocity of sound × time
1/2 ×9.8 × t1² = 343 × (6.25 - t1 )
t1 = 5.77376 sec
so height = 1/2 ×g ×t²
height = 1/2 ×9.8 × (5.773)²
height = 163.30 m
c.charge due to the reaction process between the two
The answer is 0.025J.
W=1/2*k*x^2
W=1/2*20*0.050^2
W=0.025J