(1 pt) An aricraft carrier has a mass of 8.9 x10^7kg and can cruise at 55 km/h.

a) What is the work required to stop it?
J

b) What constant force could stop it in 6 km?
N

a) If only half of the work required to stop it is done, what is its final speed?
km/h

is this from a webwork?

i d not know

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i have the same quickly

plz we neeeeeeeeeeeeeeeed help

a) convert km/h to m/s, use 1/2mv^2

b) divide a) by (the distance in meters)
c) divide the answer in a) in half and isolate v

for c) you can just directly calculate 1/2mv^2 without converting to m/s, divide the energy and then isolate v

Someone help for c idont get it

To answer these questions, we need to use the concepts of work, force, and energy.

a) What is the work required to stop the aircraft carrier?

The work done on an object can be calculated using the equation:
Work = Force x Distance x cosθ

In this case, we want to stop the aircraft carrier, so the force applied is in the opposite direction to its initial motion. Therefore, the angle (θ) between the force and the displacement is 180 degrees.

Since the question only asks for work, we assume no change in elevation, so work becomes the product of force and distance.

Given:
Mass of the aircraft carrier (m) = 8.9 x 10^7 kg
Velocity of the aircraft carrier (v) = 55 km/h = 15.28 m/s (converted from km/h to m/s)

We know that Work = Force x Distance, where Distance is the distance covered while bringing the aircraft carrier to rest.
Since the velocity is constant, the work required to stop it would be equal to the kinetic energy of the carrier.

Kinetic Energy (K.E.) = 1/2 × Mass × Velocity^2

Substituting the given values:
K.E. = 1/2 × (8.9 x 10^7 kg) × (15.28 m/s)^2

Simplifying the equation, we get:
K.E. = 1.29 x 10^10 Joules

Therefore, the work required to stop the aircraft carrier is 1.29 x 10^10 Joules.

b) What constant force could stop it in 6 km?

We can now calculate the average force required to stop the aircraft carrier over a given distance.
The work done to stop the aircraft carrier is the same as the work done by the force applied over that distance.

Given:
Distance (d) = 6 km = 6000 m (converted from km to m)

Using the equation:
Work = force x distance
Force = Work / distance

Substituting the known values:
Force = (1.29 x 10^10 J) / (6000 m)

Simplifying the equation, we get:
Force = 2.15 x 10^6 Newtons

Therefore, the constant force required to stop the aircraft carrier in 6 km is 2.15 x 10^6 Newtons.

a) If only half of the work required to stop it is done, what is its final speed?

When only half the work is done, we can say that half the kinetic energy of the aircraft carrier has been removed. Thus, the remaining half would represent the final kinetic energy.

Let's call the final speed of the aircraft carrier v_f.

Since kinetic energy is directly proportional to the square of velocity, we can write the equation as:

(1/2) × Mass × Initial Velocity^2 = (1/2) × Mass × Final Velocity^2

Substituting the known values:
(1/2) × (8.9 x 10^7 kg) × (15.28 m/s)^2 = (1/2) × (8.9 x 10^7 kg) × (v_f^2)

Simplifying the equation, we get:
v_f^2 = (1/2) × (15.28 m/s)^2
v_f^2 = 116.9392 m^2/s^2

Taking the square root of both sides, we get:
v_f = √(116.9392 m^2/s^2)

Converting the velocity to km/h:
v_f = √(116.9392) x 3600 km/h
v_f ≈ 61.25 km/h (rounded to two decimal places)

Therefore, the final speed of the aircraft carrier, if only half of the work required to stop it is done, is approximately 61.25 km/h.