Answer: D. 0.57
Explanation:
The formula to calculate the eccentricity of an ellipse is (assuming the moon's orbit in the shape of an ellipse):
Where:
is the apoapsis (the longest distance between the moon and its planet)
is the periapsis (the shortest distance between the moon and its planet)
Then:
This is the moon's orbital eccentricity
Answer:
Explanation:
For this exercise we must use the principle of conservation of energy
starting point. The proton very far from the nucleus
Em₀ = K = ½ m v²
final point. The point where the proton is stopped (v = 0)
Em_f = U = q V
where the potential is
V = k Ze / r²
Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching
Energy is conserved
Em₀ = Em_f
½ m v² = e ()
with this expression we can find the closest approach distance (r)
Answer:
The total time for the race is 11.6 seconds
Explanation:
The parameters given are;
Total distance ran by Willie = 100.0 m
Initial acceleration = 2.00m/s²
Top speed reached with initial acceleration = 12.0 m/s
Point where Willie start to fade and decelerate = 16.0 m from the finish line
Speed with which Willie crosses the finish line = 8.00 m/s
The time and distance covered with the initial acceleration are found using the following equations of motion;
v = u₀ + a·t
v² = u₀² + 2·a·s
Where:
v = Final velocity reached with the initial acceleration = 12.0 m/s
u₀ = Initial velocity at the start of the race = 0 m/s
t = Time during acceleration
a = Initial acceleration = 2.00 m/s²
s = Distance covered during the period of initial acceleration
From, v = u₀ + a·t, we have;
12 = 0 + 2×t
t = 12/2 = 6 seconds
From, v² = u₀² + 2·a·s, we have;
12² = 0² + 2×2×s
144 = 4×s
s = 144/4 =36 meters
Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;
Distance covered at top speed = 100 - 36 - 16 = 48 meters
Time, of running at top speed = Distance/velocity = 48/12 = 4 seconds
The deceleration from top speed to crossing the line is found as follows;
v₁² = u₁² + 2·a₁·s₁
Where:
u₁ = v = 12 m/s
v₁ = The speed with which Willie crosses the line = 8.00 m/s
s₁ = Distance covered during decelerating = 16.0 m
a₁ = Deceleration
From which we have;
8² = 12² + 2 × a × 16
64 = 144 + 32·a
64 - 144 = 32·a
32·a = -80
a = -80/32 = -2.5 m/s²
From, v₁ = u₁ + a₁·t₁
Where:
t₁ = Time of deceleration
We have;
8 = 12 + (-2.5)·t₁
t₁ = (8 - 12)/(-2.5) = 1.6 seconds
The total time = t + + t₁ =6 + 4 + 1.6 = 11.6 seconds.