Answer:
a) 0.3 m
b) r = 0.45 m
Explanation:
given,
q₁ = 0.44 n C and q₂ = 11.0 n C
assume the distance be r from q₁ where the electric field is zero.
distance of point from q₂ be equal to 1.8 -r
now,
E₁ = E₂
1.8 = 6 r
r = 0.3 m
<h3>b) zero when one charge is negative.</h3>let us assume q₁ be negative so, distance from q₁ be r
from charge q₂ the distance of the point be 1.8 +r
now,
E₁ = E₂
1.8 =4 r
r = 0.45 m
Answer:
x = 6.94 m
Explanation:
For this exercise we can find the speed at the bottom of the ramp using energy conservation
Starting point. Higher
Em₀ = K + U = ½ m v₀² + m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
Let's calculate
v = √(1.23² + 2 9.8 1.69)
v = 5.89 m / s
In the horizontal part we can use the relationship between work and the variation of kinetic energy
W = ΔK
-fr x = 0- ½ m v²
Newton's second law
N- W = 0
The equation for the friction is
fr = μ N
fr = μ m g
We replace
μ m g x = ½ m v²
x = v² / 2μ g
Let's calculate
x = 5.89² / (2 0.255 9.8)
x = 6.94 m