The definition of the Eclipse of the Sun says that it will form the formation of an eclipse of the sun when the moon and the sun are in exactly the same direction as seen from the earth. In such a way that the moon passes between the spectator in the earth and the sun.
We will start considering that both speed and orbit are independent factors, guarantors of the generation of an eclipse. The Moon's Orbit is inclined 5,145 ° with respect to the ecliptic plane. The existence of a solar eclipse occurs only when the moon passes through that ecliptic plane at the same time that it is closest to the sun. If you go through that plane twice, the sun will not be aligned to the 'first time', but it will be in the second time. Therefore the frequency of solar eclipses would remain the same as the frequency of alignment of the Moon, the sun and the earth will remain the same.
Therefore the correct answer is C.
Answer:
M = 0.31 kg
Explanation:
This exercise must be done in parts, let's start by finding the speed of the set arrow plus apple, for this we define a system formed by the arrow and the apple, therefore the forces during the collision are internal and the moment is conserved
let's use m for the mass of the arrow with velocity v₁ = 20.4 m / s and M for the mass of the apple
initial instant. Just before the crash
p₀ = m v₁ + M 0
instant fianl. Right after the crash
p_f = (m + M) v
p₀ = p_f
m v₁ = (m + M) v
v = (1)
now we can work the arrow plus apple set when it leaves the child's head with horizontal speed and reaches the floor at x = 8 m. We can use kinematics to find the velocity of the set
x = v t
y = y₀ + t - ½ g t²
when it reaches the ground, its height is y = 0 and as it comes out horizontally,
0 = h - ½ g t²
t² = 2h / g
For the solution of the exercise, the height of the child must be known, suppose that h = 1 m
t =
t = 0.452 s
let's find the initial velocity
v = v / t
v = 8 / 0.452
v = 17.7 m / s
From equation 1
v = m / (m + M) v₁
m + M =
M = m + m \ \frac{v_1}{v}
we calculate
M = 0.144 + 0.144
M = 0.31 kg