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<span>We calculate the electric field as follows:
r = </span>√<span>(3)/6 x 19 cm = .05484 m
The angle for the triangle would be 30 on each side.
tan(30) = r/(L/2)
E' = kQ/{r*sqrt[(L/2)^2 + r^2]} = (8.99e9 x 15e-9) / {.05484 * sqrt[(.19/2)^2 + .003]}
</span>E' <span>= 22413 N/C
The value above is the electric field strength for a single rod at the center.
|E'| = 22413 N/C
E = 2|E'|sin(30) + |E'| = 49000 N/C</span>
Answer:
x₁ = 15 m, x₂ = 12 m
, x_total = 27m, v₁ = 5 m / s
, v₂ = - 3 m / s
Explanation:
In this exercise we will use the kinematics of uniform motion
v = d / t
let's apply this equation for the first move
v₁ = Δx / t = (x₂-x₀) / t
v₁ = (12- (-3)) / 3
v₁ = 5 m / s
the distance traveled is x₁ = 15 m
Now let's analyze the return movement
v₂ = Δx / dt
v₂ = (0 - 12) / 4
v₂ = - 3 m / s
The negative sign indicates that the vehicle is moving to the left
the distance traveled is x₂ = 12 m
The total dystonia is
x_total = x₁ + x₂
x_total = 15 +12
x_total = 27m
In the attached we have the graphics of the movement
Answer:
θ = 28°
Explanation:
For this exercise We will use the second law and Newton, let's set a System of horizontal and vertical.
X axis
Fₓ = m a
Nₓ = m a
Where the acceleration is centripetal
a = v² / r
The only force that we must decompose is normal, let's use trigonometry
sin θ = Nₓ / N
cos θ =
/ N
Nₓ = N sin θ
= N cos θ
Let's replace
N sin θ = m v² / r
Y Axis
- W = 0
N cos θ = mg
Let's divide the two equations of Newton's second law
Sin θ / cos θ = v² / g r
tan θ = v² / g r
θ = tan⁻¹ (v² / g r)
We reduce the speed to the SI system
v = 61 km / h (1000 m / 1 km) (1h / 3600 s) = 16.94 m / s
Let's calculate
θ = tan⁻¹ (16.94 2 / (9.8 55.1)
θ = tan⁻¹ (0.5317)
θ = 28°