According to the theory of plate tectonics, the lithosphere is separated into sections that are called tectonic plates. These pl
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Answer:
1.28 m
Explanation:
As shown in the diagram attached,
According to the principle of moment,
For a body at equilibrium,
Sum of clockwise moment = sum of anticlockwise moment.
Taking moment about the pivot,
W₁(1.6)+W(0.133) = W₂(x)............... Equation 1
Where W₁ = Weight of the first child, Wₓ = Weight of the seesaw, W₂ = weight of the second child, x = distance of the second child from the pivot.
But,
W = mg
Where g = 9.8 m/s², m = mass of the body
Therefore,
W₁ = 26×9.8 = 254.8 N,
Wₓ = 18×9.8 = 176.4 N
W₂ = 34.4×9.8 = 337.12 N
Substitute these values into equation 1
(254.8×1.6)+(176.4×0.133) = 337.12(x)
407.68+23.4612 = 337.12x
337.12x = 431.1412
x = 431.1412/337.12
x = 1.2789
x ≈ 1.28 m
Answer:
(A) 0.279N at angle 38.02°
(B) 0.701N
(C) 14.19°
Explanation:
(A) The net force on q3 is given as:
F = Fxi + Fyj
Fx is the x component of the force
Fy is the y component of the force
Fx = -F(1, 3)cos(90 - x) + F(2, 3)cos0
Fy = -F(2, 3)cosx - F(2, 3)cos90 = -F(2, 3)cosx
First let us find y and angle x from the diagram.
Using Pythagoras theorem,
y² = 0.3² + 0.4²
y² = 0.25
y = 0.5m
Using SOHCAHTOA to find x,
sinx = 0.4/0.5
x = 53.13°
Electrostatic force, F is given as:
F = kqQ/r²
Where k = Coulumbs constant
F(1,3) = (k*q1*q3) / r²
F(1, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.5²)
F(1, 3) = 0.288N
F(2,3) = (k*q2*q3) / r²
F(2, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)
F(2, 3) = 0.45N
Therefore,
Fx = -0.288cos36.87 + 0.45
Fx = 0.22N
Fy = 0.288cos53.13
Fy = 0.172N
=> F = 0.22i + 0.172j
The magnitude of the force will be
F(mag) = √(0.22² + 0.172²)
F(mag) = 0.279N
The direction of the force makes will be
tanθ = Fy/Fx
tanθ = 0.172/0.22 = 0.781
θ = 38.02° to the x axis.
(B) q2 = - 2.0 * 10^(-6)
This implies that:
F(2,3) = (k*q2*q3) / r²
F(2, 3) = (9 * 10^9 * -2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)
F(2, 3) = -0.45N
Therefore,
Fx = -0.288cos36.87 - 0.45
Fx = -0.68N
Fy = 0.172N
=> F = - 0.68i + 0.172j
The magnitude of the force will be
F(mag) = √((-0.68)² + 0.172²)
F(mag) = 0.701N
(C) The direction of the force makes will be
tanθ = 0.172/0.68
θ = 14.19° to the x axis
