Two for are at an angle at 120 ,the bigger forces is 40N and the resultant is perpendicular to the smaller one. find the smaller
1 answer:
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Answer:
29.4 uN
Explanation:
The electric force between two charges can be calculated using Coulomb's Law. According to this law the force between two point charges is given as:
where k is a proportionality constant known as the Coulomb's law constant. Its value is Nm²/C²
r = distance between charges = 70 cm = 0.7 m
q1 = q2 = 4nC = C
The negative sign indicates that the charges are negative. In the formula we will only use the magnitude of the charges.
Using these values in the formula, we get:
Therefore, the magnitude of repulsive force between the given charges will be 29.4 uN
Answer:
a) a = 3.06 10¹⁵ m / s , b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N
Explanation:
This exercise will average solve using the moment relationship.
a ) let's use the relationship between momentum and momentum
I = ∫ F dt = Δp
F t = m - m v₀
F = m (v_{f} -v₀o) / t
in the exercise indicates that the speed module is the same, but in the opposite direction
F = m (-2v) / t
if we use Newton's second law
F = m a
we substitute
- 2 mv / t = m a
a = - 2 v / t
let's calculate
a = - 2 4.59 10²/3 10⁻¹³
a = 3.06 10¹⁵ m / s
b) F= m a
F= 4.68 10⁻²⁶ 3.06 10¹⁵
F= 1.43 10⁻¹⁰ N
c) if we hit the wall for 1015 each exerts a force F
F_total = n F
F_total = n m a
F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵
F_total = 14.32 10⁻²⁶ N