Answer:
say a lighter car is going 80mph and a heavier car is going the same speed who do you think will have more damage i will say the heavier car
Explanation:
Answer:
(a) 1/60
(b) Yes, it was unusual.
Explanation:
Number that went to the right field = 19
Number that went to the right center field = 18
Number that went to the center field = 12
Number that went to the left center field = 10
Number that went to the left field = 1
Total number of home runs = 60
(a) The probability that a randomly selected home run was hit to left field is
P(left field) = 1/60
(b) It was unusual because whenever a player hit a home run, it most times went to the right field,which is why the right field had the highest probability, 19/60.
On the other hand, it went to the left field the least number of times, which is why the left field had the least probability, 1/60.
So, it was unusual.
Answer:
Net force, F = 7 N
Explanation:
It is given that,
Initial kinetic energy of the car, 
Final kinetic energy of the car, 
Distance, d = 1 m
We need to find the average net force on the car during this interval. It is given by using the work energy theorem as :


Also, W = F.d d = distance and F = net force



F = 7 Newton
So, the average net force on the car during this interval is 7 newton. Hence, this is the required solution.
Answer:
f´ = f (v ±v₀) / (v ⁻₊ v_{s})
Explanation:
This is a doppler effect exercise that is the change of frequency by the relative speed of the emitter and receiver of sound
For the first part, the sound comes out of the sounding device, which is generally removed, so the speed of the emitter is vs = 0 and the receiver (the whale) is moving, so the equation is
f ”= f ((v ± v₀) / v)
where the sign plus (+) corresponds to the observer moving towards the source that is at rest, f is the frequency emitted by the probe and f” the frequency received by the whale.
In the second part, the waves are reflected by the whale, which becomes the emitter and the receiver that is the probe remains at rest, so the expression is
d´ = d ”(v / (v ⁻₊
))
where the least succor for the source moving towards the observer,
Therefore, the frequency received by the probe (f´) is
f´ = f ((v ±-v₀) / v) (v / v⁻₊ v_{s}))
f´ = f (v ±v₀) / (v ⁻₊ v_{s})