To solve this problem we will apply the concept given by Pythagoras in the description of the lengths of the legs of a rectangular triangle and if equality against the square of the hypotenuse, that is
Here,
a, b = Legs of a triangle
c = Hypotenuse
According to the attached chart then we would have to
Substituting the given the lengths into the Pythagorean Theorem.
Therefore the distance x is 2830km.
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Answer:
Rmax = 3.4 10⁶ m
Explanation:
For this exercise we will use the concept of energy
Initial. On the surface of the luma
Em₀ = K + U
Em₀ = ½ m v² - G m M / R_moon
Final. At the furthest point
Emf = U
Emf = - g m M / R_max
Em₀ = Emf
½ m v² - G m M / R_moon = - G m M / R_max
½ v² + G M (-1 / R_moon + 1 / R_max) = 0 (1)
Let's use the fact that tells us that the speed of the rocket is equal to the speed of a satellite that rotates around the moon near the surface, let's use Newton's second law
F = m a
Acceleration is centripetal
a = v² / r
r= R_moon
G m M / R_moon² = m v² / R_mon
G M / R_moon = v²
We substitute in 1
½ G M / R_moon + G M (1 / R_max - 1 / R_moon) = 0
1 / R_max = 1 / R_moon (1- ½)
R_max = R_moon 2
Rmax = 2 1700 103
Rmax = 3.4 10⁶ m
<span>Ohm's law deals with the relation between
voltage and current in an ideal conductor. It states that: Potential difference
across a conductor is proportional to the current that pass through it. It is
expressed as V=IR. The correct answer from the choices listed above is option A. The resistance has increased. </span>
Within Gravitational Force without obstruction.