Answer:
The average acceleration is 8.06 m/s².
Explanation:
It is given that,
Initial speed of the jet, u = 120 mph = 176 ft/s
Final velocity of the jet, v = 30 mph = 44 ft/s
Distance, d = 1800 ft
We need to find the average acceleration required of the airplane during braking. It can be calculated using third law of motion as :

a = acceleration



So, the average acceleration required of the airplane during braking is -8.06 ft/s². Hence, this is the required solution.
Answer:
a. 2.645 * 10^7 m/s^2
b. 2.645 * 10^4 N
Explanation:
Parameters given:
Velocity of rod = 2010m/s
Length of rod = 15.3cm = 0.153m
Mass of object placed at the end of the rod = 1g = 0.001kg
a. Centripetal acceleration is given as:
a = (v*v)/r
Where v = velocity
r = radius of curvature.
The radius of curvature in this case is equal to the length of the rod, since the rod makes the circular path of the motion.
Hence, centripetal acceleration at the end of the rod:
a = (2010*2010)/(0.153)
a = 26432156.86 m/s^2 = 2.64 * 10^7 m/s
b. The force needed to hold the object at the end of the rod is equal to the centripetal force at the end of the rod. Centripetal force is given as:
F = ma = (m*v*v)/r
Where a = centripetal acceleration
F = 0.001 * 2.64 * 10^7
F = 2.64 * 10^4N
Do you have the answer for this question? so i can provide u solution more effectively
It is c.
Hoped this helped.
~Bob Ross