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lbvjy [14]
3 years ago
14

What is the netforce?

Physics
2 answers:
Kipish [7]3 years ago
3 0

Answer:

6 Newtons to the left.

Explanation:

We can convert  this into a generic algebra equation by giving directions positive and negative values.

The 6 will be positive, and the 10 and 2 will be negative.

Add 10 and 2 to have 12.

6-12 = -6.

Therefore you have 6 newtons to the left (negative).

777dan777 [17]3 years ago
3 0
Take the force towards the right as positive (+6N) and towards the left as negative (-12N)
Add both as net force is the sum of all forces acting on an object
6N+(-12N)= -6N
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Adjacent antinodes of a standing wave on a string are 15.0 cmapart. A particle at an antinode oscillates in simpleharmonic motio
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Answer:

a)   D = 15 cm , b)  λ = 30.0 cm , c) 0.0850 cm, d) v = 400 cm / s , e)    v = 80.43 cm / s , v = -80.43 cm / s, f)   D₂= 7.5 cm

Explanation:

Standing waves form when two waves of the same frequency travel in opposite directions,

A) in the waves the distance of the nodes and antinode are the same since the wavelength is constant

          D = 15 cm

B) the wavelength is the distance for which the wave repeats itself, in the case of a standing wave, the distance between two nodes is lamita of the wavelength.

        D = λ / 2

         λ=  15 2

         λ = 30.0 cm

C) the amplitude of each wave is 0.0850 cm, the amplitude of the standing wave is double A = 0.17 cm

D) Let's use the speed ratio

        v =  λ f

        f = 1 / T

        v =  λ / T

       v = 30.0 /0.0750

       v = 400 cm / s

E) the transverse speeds are the speed of the oscillatory movement

       y = A cos (wt)

 

      w = 2π f = 2π / T

      w = 2π / 0.0750

      w = 83.78 rad / s

     

      y = 0.850 cos (83.78 t)

Speed ​​is

      v = dy / dt

      v = -A w cost wt

      v = - 0.850 83.78 cos (83.78 t)

      v = -80.43 cos (83.78 t)

The maximum speed when the cosine values ±1

       v = 80.43 cm / s

       v = -80.43 cm / s

F) if we draw a drawing, the distance between two nodes is half the wavelength, at the distance between an antinode synod is half this, it occupies a quarter of the wavelength

      D₂ = ¼  λ

      D₂ = 30.0/4

      D₂= 7.5 cm

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