Answer:
Explanation:
Force due to charges 1.75 and 5 nC is given below
F =K Q₁Q₂ / d²
F₁ = ![\frac{9\times10^9\times1.75\times10^{-9}\times5\times 10^{-9}}{(2.7\times10^{-2})^2}](https://tex.z-dn.net/?f=%5Cfrac%7B9%5Ctimes10%5E9%5Ctimes1.75%5Ctimes10%5E%7B-9%7D%5Ctimes5%5Ctimes%2010%5E%7B-9%7D%7D%7B%282.7%5Ctimes10%5E%7B-2%7D%29%5E2%7D)
F₁ = 10.8 X 10⁻⁵ N . It will at in x direction.
Force due to other charge placed at origin
F₂ = ![\frac{9\times10^9\times1.75\times10^{-9}\times5\times 10^{-9}}{22.5\times10^{-4}}](https://tex.z-dn.net/?f=%5Cfrac%7B9%5Ctimes10%5E9%5Ctimes1.75%5Ctimes10%5E%7B-9%7D%5Ctimes5%5Ctimes%2010%5E%7B-9%7D%7D%7B22.5%5Ctimes10%5E%7B-4%7D%7D)
F₂ = 3.5 x 10⁻⁵ N.
Its x component
= F₂ Cos θ
= 3.5 x 10⁻⁵ x 3.9/ 4.74
= 2.88 x 10⁻⁵ N
Its y component
F₂ sin θ
= 3.5 x 10⁻⁵ x 2.7/4.743
= 1.99 x 10⁻⁵ N
Total x component
= 10.8 X 10⁻⁵ +2.88 x 10⁻⁵
= 13.68 x 10⁻⁵ N.
Magnitude of total force F
F² = (13.68 x 10⁻⁵)² + (1.99 x 10⁻⁵ )²
F = 13.82 X 10⁻⁵ N
Direction θ with x axis .
Tanθ = 1.99/ 13.68
θ = 8 °
Answer:
Final Velocity = 4.9 m/s
Explanation:
We are given;. Initial velocity; u = 2 m/s
Constant Acceleration; a = 0.1 m/s²
Distance; s = 100 m
To find the final velocity(v), we will use one of Newton's equations of motion;
v² = u² + 2as
Plugging in the relevant values to give;
v² = 2² + 2(0.1 × 100)
v² = 4 + 20
v² = 24
v = √24
v = 4.9 m/s