Physics, 03.03.2020 04:28 anna4060

8) When an F-14 airplane takes off an aircraft carrier it is literally catapulted of the flight deck.
The plane's final speed at take-off is 68.2 m/s. The F-14 starts from rest. The plane accelerates
in 2 seconds and has a mass of 29 545 kg. What is the total force that gets the F-14 in the air? 9)A sports car accelerates from 0 to 27 m/s, 60 mph, in 6.3 seconds. The car exerts a force of
4106 N. What is the mass of the car?
10)A boy can accelerate at 1.00 m/s? over a short distance. If the boy were to take an energy
pill and suddenly have the ability to accelerate at 5.6 m/s“, then how would his new energy-pill-
force compare to his earlier force? If the boy's earlier force was 45 N, what is the size of his
energy-pill-force?
12) A cartoon plane with four engines can accelerate at 8.9 m/s- when one engine is running.
What is the acceleration of the plane if all four engines are running and each produces the same
force?

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Answer from: elliehemms6324

8) $1.0\cdot 10^6 N$

9) 958 kg

10) F_2=5.6 F_1, F_2=252 N

11) 35.6 m/s^2

Explanation:

The total force exerted on an object is equal to the rate of change of momentum of the object; mathematically:

$F=\frac{\Delta p}{\Delta t}$

where

F is the force

$\Delta p$ is the change in momentum

$\Delta t$ is the time elapsed

The change in momentum can be written as

$\Delta p = m(v-u)$

where

m is the mass

v is the final velocity

u is the initial velocity

For the plane in this problem:

u = 0 (it starts from rest)

v = 68.2 m/s

$\Delta t = 2 s$

m = 29545 kg

Therefore, the force exerted on the F-14 is:

$F=\frac{(29545)(68.2-0)}{2}=1.0\cdot 10^6 N$

As in the previous problem, the force exerted on the car can be written as

$F=\frac{\Delta p}{\Delta t}=\frac{m(v-u)}{\Delta t}$

where in this problem, we have:

u = 0 is the initial velocity of the car

v = 27 m/s is the final velocity of the car

$\Delta t = 6.3 s$ is the time elapsed during the acceleration of the car

F = 4106 N is the force exerted on the car

Re-arranging the equation for m, we can find the mass of the car:

$m=\frac{F\Delta t}{v-u}=\frac{(4106)(6.3)}{27-0}=958 kg$

So, the mass of the car is 958 kg.

10)

The relationship between force exerted on a body and acceleration of the body is given by Newton's second law of motion:

F=ma

where

F is the force

m is the mass

a is the acceleration

At the beginning, with a certain force F_1 , the acceleration of the boy is

a_1=1.00 m/s^2

So we have

F_1=ma_1 (1)

Later, after the boy has taken the pill, the force is F_2 , while the acceleration is now

a_2=5.6 m/s^2

So we have

F_2=ma_2 (2)

To do a comparison between the two forces, we divide (2) by (1), and we get:

$\frac{F_2}{F_1}=\frac{a_2}{a_1}=\frac{5.6}{1}=5.6$

So,

F_2=5.6 F_1

Which means that the force in the 2nd case is 5.6 times larger than the force in the 1st case.

We are also told that the boy's earlier force is

F_1=45 N

Therefore, the force in the 2nd case is:

F_2=5.6(45)=252 N

12)

For this problem we can use again Newton's second law of motion:

F=ma

where

F is the force

m is the mass

a is the acceleration

At the beginning, with one engine only running, the acceleration of the plane is

a=8.9 m/s^2

And we can write it as

$a=\frac{F}{m}$

where F is the force exerted by 1 engine only.

Later, when all 4 engines are running, the new force will be 4 times:

F' = 4F

Therefore, the new acceleration will be:

$a'=\frac{F'}{m}=\frac{4F}{m}=4(\frac{F}{m})=4a$

So, 4 times the previous acceleration; and therefore:

a'=4a=4(8.9)=35.6 m/s^2

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8) When an F-14 airplane takes off an aircraft carrier it is literally catapulted of the flight deck...

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