Answer:
When the muscle is completely contract.
Explanation:
Remember the that maximum force of a muscle is when is completely contract. A characteristic of a muscle is that can contract and can relax in the opposite direction. In this way, when all the microfibers of the muscle are join together (they are contract) is when the maximum tissue force is applied.
With exercise the fibers of the muscles can grow or reproduce to strength the muscle.
r₁ = distance of point A from charge q₁ = 0.13 m
r₂ = distance of point A from charge q₂ = 0.24 m
r₃ = distance of point A from charge q₃ = 0.13 m
Electric field by charge q₁ at A is given as
E₁ = k q₁ /r₁² = (9 x 10⁹) (2.30 x 10⁻¹²)/(0.13)² = 1.225 N/C towards right
Electric field by charge q₂ at A is given as
E₂ = k q₂ /r₂² = (9 x 10⁹) (4.50 x 10⁻¹²)/(0.24)² = 0.703 N/C towards left
Since the electric field in left direction is smaller, hence the electric field by the third charge must be in left direction
Electric field at A will be zero when
E₁ = E₂ + E₃
1.225 = 0.703 + E₃
E₃ = 0.522 N/C
Electric field by charge "q₃" is given as
E₃ = k q₃ /r₃²
0.522 = (9 x 10⁹) q₃/(0.13)²
q₃ = 0.980 x 10⁻¹² C = 0.980 pC
Answer:F=1.7802
Explanation:
Since we've been given the mass to be .18kg,we are asked to find the Force of which the formulae is
F=ma where f-force,m-mass and a-acceleration due to gravity
So we can just substitute
F-?.m=.18 and a9.89
F=.18×9.89
F=1.7802N