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mixer [17]
3 years ago
7

If the standard kilogram bar kept in Paris were subjected to a net force of 1 newton, what acceleration would it have as a resul

t?
Physics
2 answers:
sdas [7]3 years ago
7 0

Answer:

It would have an acceleration of 1\frac{m}{s^{2}}

Explanation:

Let's start defining the newton unit (Newton : N)

The N definition is

N=Kg.(\frac{m}{s^{2}})

Where Kg is kilogram and \frac{m}{s^{2}} is an acceleration unit.

We can think the definition of newton as :

Given an object with a mass of 1 Kg subjected to a force of 1 N the acceleration that it experiments is 1\frac{m}{s^{2}}

As a result, the standard  kilogram bar kept in Paris will experiment an acceleration of 1\frac{m}{s^{2}} If we subject it to a net force of 1 N.

Klio2033 [76]3 years ago
3 0
1 m/s^2
Because of
F=ma
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Light is shone on a diffraction grating
Pani-rosa [81]

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in this case they indicate the distance between slits, the angle and the order of diffraction

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let's calculate

         λ = 1.00 10⁻⁶ sin 74.6 / 2

         λ = 4.82048 10⁻⁷ m

Let's reduce to nm

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1. Two forces act on a box as follows: F1 = 100 N at 01 = 170° and F2 = 75 N
lilavasa [31]

Answer:

a)  F = 64.30 N,  b) θ = 121.4º

Explanation:

Forces are vector quantities so one of the best methods to add them is to decompose each force and add the components

let's use trigonometry

Force F1

          sin 170 = F_{1y} / F₁

          cos 170 = F₁ₓ / F₁

          F_{1y} = F₁ sin 170

          F₁ₓ = F₁ cos 170

          F_{1y} = 100 sin 170 = 17.36 N

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Force F2

          sin 30 = F_{2y} / F₂

          cos 30 = F₂ₓ / F₂

          F_{2y} = F₂ sin 30

          F₂ₓ = F₂ cos 30

          F_{2y} = 75 sin 30 = 37.5 N

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the resultant force is

X axis

          Fₓ = F₁ₓ + F₂ₓ

          Fₓ = -98.48 +64.95

          Fₓ = -33.53 N

Y axis

         F_y = F_{1y} + F_{2y}

         F_y = 17.36 + 37.5

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a) the magnitude of the resultant vector

let's use Pythagoras' theorem

         F = Ra Fx ^ 2 + Fy²

         F = Ra 33.53² + 54.86²

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b) the direction of the resultant

let's use trigonometry

        tan θ’= F_y / Fₓ

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this angle is in the second quadrant

The angle measured from the positive side of the x-axis is

        θ = 180 -θ'

        θ = 180- 58.6

        θ = 121.4º

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