Scenario
1 answer:
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
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1) Mechanical energy is conserved in all the 4 situations
2) True
3) The potential energy of the block is 490 J
Explanation:
1)
Mechanical energy is the sum of potential energy (PE) and kinetic energy (KE) of a body:
According to the law of conservation of energy, in absence of non-conservative forces (such as air resistance, friction...), the mechanical energy of a body is always conserved.
This means that the mechanical energy is conserved in all the situations described. More specifically:
- Child on a swing : there is a continuous conversion between kinetic energy and gravitational potential energy
- Pendulum : there is a continuous conversion between kinetic energy and gravitational potential energy
Bow and Arrow : there is a conversion of energy from elastic potential (when the bow is stretched) to kinetic energy (when the arrow is shot)
Roller Coaster: there is a continuous conversion between kinetic energy and gravitational potential energy
2)
As we said, the mechanical energy of the object falling down at any point of the fall is
where KE is the kinetic energy and PE is the potential energy. The value of E is constant, since the mechanical energy is conserved.
The potential energy is given by
where m is the mass of the object, g is the acceleration of gravity, h is the height of the object above the ground.
As the object falls, its height decreases, and therefore, the potential energy PE also decreases. But we said that E must remain constant: therefore, if PE decreases, this means that KE increases, therefore as the object falls, the potential energy is converted into kinetic energy (in fact, the speed of the object increases). So the statement is true.
3)
The potential energy of an object is given by the equation
where
m is the mass of the object
is the acceleration of gravity
h is the height of the object relative to the ground
For the block in this problem, we have:
m = 10 kg
h = 5 m
Substituting, we find its potential energy:
Learn more about potential energy and kinetic energy:
#LearnwithBrainly
a) E = 0
b)
Explanation:
a)
We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:
where
E is the electric field
q is the charge contained by the Gaussian surface
is the vacuum permittivity
Here we want to find the electric field at a distance of
r = 12 cm = 0.12 m
Here we are between the inner radius and the outer radius of the shell:
However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.
This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:
q = 0
Therefore, the magnitude of the electric field is also zero:
E = 0
b)
Here we want to find the magnitude of the electric field at a distance of
r = 20 cm = 0.20 m
from the centre of the shell.
Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.
Therefore, it is given by:
where in this problem:
is the charge on the shell
is the distance from the centre of the shell
Substituting, we find: