(a) No, because the mechanical energy is not conserved
Explanation:
The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:
(1)
However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.
Therefore, eq. (1) can be rewritten as

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane (
) and part is lost because of the air resistance (
).
(b) 77.8 m/s
First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

Now we can calculate the acceleration of the plane, by using Newton's second law:

where m is the mass of the plane.
Finally, we can calculate the final speed of the plane by using the equation:

where
is the final velocity
is the initial velocity
is the acceleration
is the distance travelled
Solving for v, we find

For simplicity, let's call vector B-A vector C Then C is
Cx = (-6.1 - 2.2)
Cy = (-2.2 - (-6.9)) Or,
Cx = -8.3 Cy = 4.7
The magnitude is found with the Pythagorean theorem
||C|| = √(-8.3² + 4.7²) = 9.538
If an experiment is conducted such that an applied force is exerted on an object, a student could use the graph to determine the net work done on the object.
The graph of the net force exerted on the object as a function of the object’s distance traveled is attached below.
- A student could use the graph to determine the net work done on the object by Calculating the area bound by the line of best fit and the horizontal axis from 0m to 5m
For more information on work done, visit
brainly.com/subject/physics
Answer:
≅50°
Explanation:
We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:
Δx=V₀t+at²/2
And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:
Δx=(V₀cosθ)t+at²/2
Now luckily we are given everything we need to solve (or you found the info before posting here):
- Δx=760 m
- V₀=87 m/s
- t=13.6 s
- a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!
With that we can plug the values in to get:



