3. You are flying 2586 miles from San Francisco to New York. An hour into the flight, you are 600
2 answers:
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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
Explanation:
The given data is as follows.
mass (m) = 170 kg, Distance (s) = 9.6 m
Height (h) = 3.3 m, Force (F) = 1400 N
First, we will calculate the work performed by her as follows.
W = Fs
=
= 13440 J
Hence, minimal work necessary to lift the refrigerator is as follows.
U = mgh
=
= 5497.8 J
Therefore, we can conclude that he performed 5497.8 J of work.
The expression of the electric flux is
Here,
Q = Total charge enclosed in the closed surface
= Permittivity due to free space
Rearranging to find the charge,
Replacing with our values we have finally
The charge enclosed by the box is 0.1684nC
The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.