WILL MARK BRAINLIEST!!!!
1 answer:
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Answer:
r = <2.640 10⁶, 1.01 10⁸, 0> m
Explanation:
For this exercise we are going to solve it for each direction separately,
we locate a fixed reference frame in space at the height of the rocket, such that the position of the rocket is
r₀ = <4, 4, 0> 10³ m
X axis
the initial velocity of the ship on this axis is v₀ₓ = 0, when it passes through the point x₀ = 4 km it ignites the rockets, experiencing a force of Fₓ = 7.0 10⁵ N for 21 s and the rockets turn off
They ask where it is after one hour t = 1 h = 3600 s
Let's apply Newton's second law
Fₓ = m aₓ
aₓ = Fₓ / m
aₓ = 7.0 10⁵/2 10⁴
aₓ = 3.5 10¹ m / s
Let's use kinematics to find the distance
for the first t₁ = 21 s the movement is accelerated
x₁ = x₀ + v₀ t₁ + ½ aₓ t₁²
x₁ = x₀ + ½ aₓ t₁²
x₁ = 4000 + ½ 35 21² = 4000 + 7717,.5
x₁ = 11717.5 m
this instant has a speed of
vₓ = v₀ₓ + aₓ t
vₓ = aₓ t ₁
vₓ = 35 21
vₓ = 735 m / s
the rest of the time there is no acceleration so it is uniform motion at this speed
t₂ = 3600 - 21
t₂ = 3576 s
vₓ = x₂ / t₂
x₂ = vₓ t₂
x₂ = 735 3576
x₂ = 2,628 10⁶ m
the total distance traveled in this direction is
x_total = x₁ + x₂
x_total = 11717.5 + 2.628 10⁶
x_total = 2,640 10⁶ m
Y axis
on this axis it is in the initial position of y₀ = 4000 m, with an initial velocity of = 28 10³ m / s and there is no force on this axis F_{y} = 0
The movement in this axis is uniform,
v_{y} = y / t
y = v_{y} t
y = 28 10³/3600
y = 1.01 10⁸ m
the total distance is
y_total = y₀ + y
y_total = 4000 + 1.01 10⁸
y_total = 1.01 10⁸ m
Z axis
the initial position is z₀ = 0, with an initial velocity of v₀ = 0 and in this axis there is no force F_{z} = 0
the movement is uniform
z = 0
the final position of the rocket after 1 h is
r = <2.640 10⁶, 1.01 10⁸, 0> m
A) -2.0 m
Look at the ray diagram attached in the picture, where:
p identifies the location of the object
q identifies the location of the image
F identifies the focus of the mirror
Each tick represents 1 m
We have
p = 6.0 m is the distance of the object from the mirror
f = -3.0 m is the focal length
From the ray diagram, we see that q has a distance of 2.0 m from the mirror, and it's on the other side of the mirror compared to the object, so
q = -2.0 m
This can also be verified by using the mirror equation:
B) Upright and virtual
As we see from the picture, the image is upright, since it has same orientation as the object.
Also, we notice that the image is on the other side of the mirror, compared to the object. For a mirror,
- An image is said to be real if it is on the same side of the object
- An image is said to be virtual if it is on the opposite side of the mirror
Therefore, this means that the image is virtual.
