Answer:
4.763 × 10⁶ N/C
Explanation:
Let E₁ be the electric field due to the 4.0 μC charge and E₂ be the electric field due to the -6.0 μC charge. At the third corner, E₁ points in the negative x direction and E₂ acts at an angle of 60 to the negative x - direction.
Resolving E₂ into horizontal and vertical components, we have
E₂cos60 as horizontal component and E₂sin60 as vertical component. E₁ has only horizontal component.
Summing the horizontal components we have
E₃ = -E₁ + (-E₂cos60) = -kq₁/r²- kq₂cos60/r²
= -k/r²(q₁ + q₂cos60)
= -k/r²(4 μC + (-6.0 μC)(1/2))
= -k/r²(4 μC - 3.0 μC)
= -k/r²(1 μC)
= -9 × 10⁹ Nm²/C²(1.0 × 10⁻⁶)/(0.10 m)²
= -9 × 10⁵ N/C
Summing the vertical components, we have
E₄ = 0 + (-E₂sin60)
= -E₂sin60
= -kq₂sin60/r²
= -k(-6.0 μC)(0.8660)/(0.10 m)²
= -9 × 10⁹ Nm²/C²(-6.0 × 10⁻⁶)(0.8660)/(0.10 m)²
= 46.77 × 10⁵ N/C
The magnitude of the resultant electric field, E is thus
E = √(E₃² + E₄²) = √[(-9 × 10⁵ N/C)² + (46.77 10⁵ N/C)²) = (√226843.29) × 10⁴
= 476.28 × 10⁴ N/C
= 4.7628 × 10⁶ N/C
≅ 4.763 × 10⁶ N/C
Structural constraint is the answer :)
If 50 identical light bulbs are connected in series across
a single power source, then the voltage across each bulb
is ( 1/50 ) of the voltage delivered by the power source.
Given:
Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.
We need to determine the maximum shear stress developed in the beam:
τ = F/A
Assuming the area of the beam is 100 m^2 with a length of 10 m.
τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2<span />
They can either cancel each other or add up to a resultant force with a certain direction and modulus.
Newton's second law states that F=m*a, where F is the resultant force, ie ΣF.
