What is the velocity of a jet that travels 785 m [W] in 2.39 s?
1 answer:
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We can apply the law of conservation of energy here. The total energy of the proton must remain constant, so the sum of the variation of electric potential energy and of kinetic energy of the proton must be zero:
which means
The variation of electric potential energy is equal to the product between the charge of the proton (q=1eV) and the potential difference (
):
Therefore, the kinetic energy gained by the proton is
<span>And since the initial kinetic energy of the proton was zero (it started from rest), then this 1000 eV corresponds to the final kinetic energy of the proton.</span>
which means
The variation of electric potential energy is equal to the product between the charge of the proton (q=1eV) and the potential difference (
Therefore, the kinetic energy gained by the proton is
<span>And since the initial kinetic energy of the proton was zero (it started from rest), then this 1000 eV corresponds to the final kinetic energy of the proton.</span>
Answer:
d= 64.1 km θ = 36.4º
Explanation:
a) In order to find the plane's straight-line distance from the starting point, we need to know the coordinates of the final and initial position of the plane, so we can find the total displacement, as the difference between the final and initial position.
If we choose to put our origin at the initial point of trajectory, we have that (x₀, y₀) = (0, 0)
In order to find the position of the plane after finishing the flight, we need to find its final coordinates (x₁, y₁).
In order to get x₁, we need to add the x-coordinate after flying 45º north of east, and the Δx after completing the flight in a direction 25º of east, that we can find applying trigonometry, as follows:
x₁ = 37.0 km * cos 45º + 28.0 km* cos 25º = 51.6 Km
Appying the same considerations for the y-coordinate, we have:
y₁ = 37.0 km * sin 45º + 28.0 km* sin 25º = 38.0 km
Now, as the initial position coincides with the origin, the distance in a straight line from this point to the origin, is just the hypotenuse of the triangle determined by the coordinates (x₁, y₁) and (0,0), as follows:
The geographic direction of the displacement vector (which coincides in magnitude with the distance we have just found), is just the angle that this distance forms with the east axis, that we can find getting the tangent of this angle as follows:
tg θ =
⇒ θ = 36.4º North of East (counterclockwise from the east axis).