Which planet moves the fastest and which planet moves slowest? a, b, c, d, e, or f. Explain
1 answer:
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Answer:
a) F_net = 6.48 10⁻¹⁸ ( ), b) x = 0.15 m
Explanation:
a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.
The electric force is given by Coulomb's law
F =
Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.
F_net= ∑ F = F₁ + F₂
let's locate a reference system in the load that is on the left side, the distances are
left side - electron r₁ = x
right side -electron r₂ = d-x
let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q
we substitute
F_net = k q Q ( )
F _net = kqQ ( )
let's substitute the values
F_net = 9 10⁹ 1.6 10⁻¹⁹ 4.50 10⁻⁹ ( )
F_net = 6.48 10⁻¹⁸ ( )
now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values of the distance (x) and the net force are given
x (m) F (N)
0.05 27.0 10-16
0.10 8.10 10-16
0.15 5.76 10-16
0.20 8.10 10-16
0.25 27.0 10-16
b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m
Answer:
a) The maximum height reached by the projectile is 558 m.
b) The projectile was 21.3 s in the air.
Explanation:
The position and velocity of the projectile at any time "t" is given by the following vectors:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
v = (v0 · cos α, v0 · sin α + g · t)
Where:
r = position vector at time "t"
x0 = initial horizontal position
v0 = initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).
v = velocity vector at time t
a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:
vy = v0 · sin α + g · t
0 = 175 m/s · sin 36.7° - 9.80 m/s² · t
- 175 m/s · sin 36.7° / - 9.80 m/s² = t
t = 10.7 s
Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).
Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.
y = y0 + v0 · t · sin α + 1/2 · g · t²
y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²
y = 558 m
The maximum height reached by the projectile is 558 m
b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.
We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²
0 = t (175 m/s · sin 36.7 - 1/2 · 9.8 m/s² · t)
0 = 175 m/s · sin 36.7 - 1/2 · 9.8 m/s² · t
- 175 m/s · sin 36.7 / -(1/2 · 9.8 m/s²) = t
t = 21.3 s
The projectile was 21.3 s in the air.
