Answer:
The deceleration is
Explanation:
From the question we are told that
The distance of the car from the crossing is
The speed is
The reaction time of the engineer is
Generally the distance covered during the reaction time is
=>
=>
Generally distance of the car from the crossing after the engineer reacts is
=>
=>
Generally from kinematic equation
Here v is the final velocity of the car which is 0 m/s
So
=>
Answer:
B. 2.77 x N
Explanation:
The required force can be calculated by:
F =
Where F is the force between the particles, K is the coulomb's constant (9 x
),
is the charge on the first particle,
is the charge on the second particle and
is the distance between the particles.
So that:
F =
=
= 2.7692 x
The force between the particles is 2.77 x N.
Answer:
E = Q / 4 π ε₀ (r-R_int) / (R_put³ -R_int³)
Explanation:
For this exercise we can use Gauss's law
Ф = ∫ E. dA = / ε₀
Where q is the charge inside the surface.
In this case the surface must be a sphere, the electric field lines and the radius of the sphere are parallel, so the scalar product is reduced to the algebraic product
∫ E dA = q_{int}/ ε₀
The area of a sphere is
A = 4π r²
- The electric field for a distance r < R_int
The charge inside is zero, so the electric field
E = 0 r <R_in t
- The field for a radio inside the shell
Let's use the concept of density
ρ = Q / V
q = ρ (4/3 π r³)
dq = ρ 4π r² dr
We substitute in the Gaussian equation
E ∫ dA = ρ 4π r² dr / ε₀
E 4π r² = ρ 4π/ε₀ r³ / 3
E = ρ / 3ε₀ r
We evaluate between the lower limit r = R_int, E = 0 and the upper limit r = r, E = E
E- 0 = ρ / 3ε₀ (r –R_int)
Density is
ρ = q / 4/3 π (R_out³ - R_int³)
Where R <r
E = Q / 4 π ε₀ (r-R_int) / (R_put³ -R_int³)