Un auto que acelera a razón de 2 m/S^2 ; Si para cierto tramo se observa que logra triplicar su rapidez en 6 s. ¿Cuál fue su rap
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Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
The potential at a distance can be represented as:
V(r) - V(0) =
V(r) - V(0) = ₀
V(r) = ₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) =
V(r) - V(0) =
V(r) =
V(r) =
V(r)
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V
They were going at a velocity 4.07m/s
<u>Explanation:</u>
Distance s =5 m
initial velocity u= 0.8 m/s
Acceleration a =1.6m/s2
We have to calculate the velocity with which they were going afterwards i.e final velocity.
Use the equation of motion
They were going with a velocity 4.07 m/s afterwards.