<span>Time needed for drive a distance 100 km at 100 km/h:
t1 =100/100
= 1 h
Time needed for drive a distance 40 km at 45 km/h:
t2 =40/45
=0.8 h
t3 =3h 15m - 1 h- 0.8 h
=1.45 h
Distance to town equals:
s3 =270-100-40
=130 km
Least speed needed to arrive in time for the interview:
v3 =130 km/1.45 h
=89.65 km/h
The answer is 89.65 km/h</span>
Answer:
Given:
Mass of elephant = 5240 kg
The initial speed of the elephant = 4.55 m/s
Mass of the rubber ball, m, = 0.15 kg
Inital speed of the rubber ball, v = 7.81 m/s
On substitution in 
= ![[\frac{2m_{1} }{m_{1} +m_{2} } ]V_{1f}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B2m_%7B1%7D%20%7D%7Bm_%7B1%7D%20%2Bm_%7B2%7D%20%7D%20%5DV_%7B1f%7D)
+ ![[\frac{m_{2}-m_{1}}{m_{1}+m_{2} } ] v_{2f}](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bm_%7B2%7D-m_%7B1%7D%7D%7Bm_%7B1%7D%2Bm_%7B2%7D%20%20%7D%20%5D%20v_%7B2f%7D)
= ](https://tex.z-dn.net/?f=%5B%5Cfrac%7B2%2A5240_%7B%7D%20%7D%7B5240_%7B%7D%20%2B0.15_%7B%7D%20%7D%20%5D%28-4.55_%7B%7D%29)
+ ![[\frac{0.15_{}-5240_{}}{5240_{}+0.15_{} } ] (7.81_{})](https://tex.z-dn.net/?f=%5B%5Cfrac%7B0.15_%7B%7D-5240_%7B%7D%7D%7B5240_%7B%7D%2B0.15_%7B%7D%20%20%7D%20%5D%20%287.81_%7B%7D%29)
a) The negatıve sign shows that the ball bounces back in the direction opposite to the incident
b) it is clear that the velocity of the ball increases and therefore it is kinetic energy
. The ball gains kinetic energy from the elephant.
It makes it easier to dig a hole because it helps pick up dirt
Is there any possible chance that at some point in your science
studies, sometime before you were given this question for your
homework, that maybe you might have encountered this formula
for the period of a simple pendulum ?
Period = (2 pi) √(length/gravity) .
If the length is 0.23 meter, and the
acceleration of gravity is 9.8 m/s²,
then the period is
= (2 pi) √(0.23/9.8)
= 0.963... second (rounded)
That's how long it takes for a simple pendulum, 23cm long,
hanging on a massless string and not swinging too far to
the side, to complete one full swing left and right.
Now, if you can figure out how many periods of 0.963 second
there are in 30 seconds, you'll have your answer. I'll leave
that part of it to you.
