Answer:
C. The bug's change in momentum is equal to the car's change in momentum.
Explanation:
As we know by Newton's 2nd law

here we have also know that when car hits the bug then force applied by wind shield on the bug is same as the force applied by the bug on the car's wind shield as per Newton's III law

so we know that

so we have

so correct answer will be
C. The bug's change in momentum is equal to the car's change in momentum.
Answer:
69.68 N
Explanation:
Work done is equal to change in kinetic energy
W = ΔK = Kf - Ki = 
W = 
where m = mass of the sprinter
vf = final velocity
vi = initial velocity
W = workdone
kf = final kinetic energy
ki = initial kinetic energy
d = distance traveled
Ftotal = total force
vf = 8m/s
vi= 2m/s
d = 25m
m = 60kg
inserting parameters to get:
W = ΔK = Kf - Ki = 



= 39.7
we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

The average acceleration can be found by dividing the final speed by the time taken to reach said point so in this case you divide 60 by 8 resulting in 7.5 which will be your answer
<span>500 km x (1 hr/88 km) = 5.68 hrs</span>
The student's shoulder supports the weight of the bag.
<h3>What is the free body diagram?</h3>
Free-body diagrams are utilized to display the relative direction and strength of all forces that are being applied to an item in a certain scenario. A unique illustration of the geometric diagrams that were covered in a previous lesson is the free-body diagram. We will make use of these graphics throughout the entire study of physics.
A university student is carrying a backpack. One strap is hanging the rucksack immobile from one shoulder.
The weight of the backpack is balanced by the shoulder of the student.
The free-body diagram is attached below.
More about the free body diagram link is given below.
brainly.com/question/24087893
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