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Mademuasel [1]
3 years ago
13

A chair of weight 85.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F =

40.0 N directed at an angle of 35.0deg below the horizontal and the chair slides along the floor.
Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.
Physics
1 answer:
blsea [12.9K]3 years ago
6 0

Answer:

 N = 107.94 N

Explanation:

For this exercise we must use Newton's second law.

Let's set a reference system with the x-axis parallel to the ground and the y-axis vertical

X axis

        Fₓ = ma

ej and

       N -F_y - W = 0

let's use trigonometry to decompose the applied force

     cos -35 = Fₓ / F

     sin -35 = F_y / F

     Fₓ = F cos -35

     F_y = F sin -35

     Fₓ = 40.0 cos -35 = 32.766 N

     F_y = 40.0 sin -35 = -22.94 N

we substitute

     N = Fy + W

     N = 22.94 + 85

     N = 107.94 N

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A. is true

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Two charged particles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged par
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Answer:

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Explanation:

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Since all charges are of the same sign, forces are repulsive

        F₁₃ = k q₁ q₃ / r₁₃²

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         r₁₃ = x₃- 0

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We substitute

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        4- 4x₃ + x₃² = 4 x₃²

        5x₃² + 4 x₃ - 4 = 0

We solve the quadratic equation

        x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2  5

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7 0
3 years ago
You are standing 2.5m directly in front of one of the two loudspeakers. They are 3.0m apart and both are playing a 686Hz tone in
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distance from speaker is 17.87 m

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given data

distance from loudspeaker = 2.5 m

distance between loudspeaker = 3.0 m

room temperature = 20c

wavelength f  = 686Hz

to find out

what distances from the speaker

solution

we know sound velocity c = 331.5  + 0.6 × 20c = 343.5

so wavelength of sound  λ = c / f  

wavelength = 343.5 /  686 = 0.5 m

when the difference in distance of speaker destructive interference will be

d = λ/2 × (2n-1)

for n = 1, 2 3 4 ..

d = 0.5/2 × (2n-1)

d = 0.250 , 0.75 , 1.25 , 1.750............   for n = 1, 2 3 .............

so

for d = 0.250

side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x1) = 0.250

0.5 x1 = 7.6875

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side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x2) = 0.75

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side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x3) = 1.250

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x4 will be negative so we stop here

so the distance from speaker here is given below

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distance 1 = 2.5 + 0.475  = 2.975 m

distance 2 = 2.5 + 3.125  = 5.625 m

distance 3 = 2.5 + 15.375 = 17.875 m

final distance from speaker is 17.87 m

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