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sdas [7]
2 years ago
5

Mike stands on a scale in an elevator. If the elevator is accelerating upwards with 4.9 m/s2, the scale reading is ____ times Mi

ke's weight.
Physics
1 answer:
Bas_tet [7]2 years ago
7 0

Answer:

F - M a      force exerted by scales on student

M a = M (9.8 + 4.9) m/s^2      upwards chosen as positive

a = 1.5 g        net acceleration of student  due to force of scales

W =M g       weight of student   (actual weight)

Wapp = M 1.5 * g      apparent weight (on scales) of student

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IP The x and y components of a vector r⃗ are rx = 16 m and ry = -8.5 m , respectively
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as it is given that

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A point charge q1=2.0μC is located on the positive y axis at y=0.30m, and an identical charge q2 is at the origin. Find the magn
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(A) 0.279N at angle 38.02°

(B) 0.701N

(C) 14.19°

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(A) The net force on q3 is given as:

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Fx is the x component of the force

Fy is the y component of the force

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Fy = -F(2, 3)cosx - F(2, 3)cos90 = -F(2, 3)cosx

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Using SOHCAHTOA to find x,

sinx = 0.4/0.5

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F = kqQ/r²

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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

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The magnitude of the force will be

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The direction of the force makes will be

tanθ = Fy/Fx

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θ = 38.02° to the x axis.

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F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * -2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

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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

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