I am starting out as a pilot. What single engine make of airplane should I get (Cheaper to bye one and train than to rent it)?
2 answers:
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Answer:
1) t = 23.26 s, x = 8527 m, 2) t = 97.145 s, v₀ = 6.4 m / s
Explanation:
1) First Scenario.
After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.
Before starting, let's reduce the magnitudes to the SI system
v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s
y = 2650 m
Let's start by looking for the time it takes for the load to reach the ground.
y = y₀ + v_{oy} t - ½ g t²
in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero
0 = y₀ + 0 - ½ g t2
t =
t = √(2 2650/9.8)
t = 23.26 s
this is the horizontal scrolling time
x = v₀ t
x = 366.67 23.26
x = 8527 m
the speed at the point of arrival is
v_y = v_{oy} - g t = 0 - gt
v_y = - 9.8 23.26
v_y = -227.95 m / s
Module and angle form
v =
v = √(366.67² + 227.95²)
v = 431.75 m / s
θ = tan⁻¹ (v_y / vₓ)
θ = tan⁻¹ (227.95 / 366.67)
θ = - 31.97º
measured clockwise from x axis
We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.
2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º
we look for the components of speed
cos θ = v₀ₓ / v₀
sin θ = v_{oy} / v₀
v₀ₓ = v₀ cos θ
v_{oy} = v₀ sin θ
we look for the time for the arrival point that has coordinates x = 0, y = 0
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + vo sin θ t - ½ g t²
0 = -50 + vo sin 50 t - ½ 9.8 t²
x = x₀ + v₀ₓ t
0 = x₀ + vo cos θ t
0 = -400 + vo cos 50 t
podemos ver que tenemos un sistema de dos ecuación con dos incógnitas
50 = 0,766 vo t – 4,9 t²
400 = 0,643 vo t
resolved
50 = 0,766 ( ) t – 4,9 t²
50 = 476,52 t – 4,9 t²
t² – 97,25 t + 10,2 = 0
we solve the quadratic equation
t = [97.25 ± ] / 2
t = 97.25 ±97.04] 2
t₁ = 97.145 s
t₂ = 0.1 s≈0
the correct time is t1 the other time is the time to the launch point,
t = 97.145 s
let's find the initial velocity
x = x₀ + v₀ cos 50 t
0 = -400 + v₀ cos 50 97.145
v₀ = 400 / 62.44
v₀ = 6.4 m / s
Answer:
The x-component of the electric field at the origin = -11.74 N/C.
The y-component of the electric field at the origin = 97.41 N/C.
Explanation:
<u>Given:</u>
- Charge on first charged particle,
- Charge on the second charged particle,
- Position of the first charge =
- Position of the second charge =
The electric field at a point due to a charge at a point
distance away is given by
where,
= Coulomb's constant, having value
= position vector of the point where the electric field is to be found with respect to the position of the charge
.
= unit vector along
The electric field at the origin due to first charge is given by
is the position vector of the origin with respect to the position of the first charge.
Assuming, are the units vectors along x and y axes respectively.
Using these values,
The electric field at the origin due to the second charge is given by
is the position vector of the origin with respect to the position of the second charge.
Using these values,
The net electric field at the origin due to both the charges is given by
Thus,
x-component of the electric field at the origin = -11.74 N/C.
y-component of the electric field at the origin = 97.41 N/C.