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MArishka [77]
4 years ago
10

hi :) is decreasing acceleration non uniform acceleration ? anyone able to explain I’m rlly confused. Thanks!

Physics
1 answer:
Genrish500 [490]4 years ago
5 0

Answer:

Yes

Explanation:

Non uniform acceleration is any acceleration that is not constant.

If you look at the graphs I have drawn in the above picture, the first two graphs shows a uniform acceleration.

The first graph is a positive acceleration, which means that the object is moving faster and faster at a constant rate. The second graph shows a deceleration, or negative acceleration, which means that the object is moving slower and slower at a constant rate.

For velocity- time graphs, acceleration can be seen by its gradient. So if the slope of the graph doesn't change, it has a uniform acceleration.

Graph 3 shows zero acceleration since the object is moving at a constant velocity (or speed). Thus, the object does not acceleration.

Graphs 4-7 shows a non uniform acceleration.

In graph 4, the object has a decreasing acceleration since the gradient of the graph is decreasing. This can be seen by the slope getting gentler and gentler.

Graph 5 shows an increasing deceleration, since the graph is getting steeper and steeper and the velocity is decreasing with time.

Graph 6: increasing acceleration

Graph 7: decreasing deceleration

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Answer:

                  = 5.241 \times 10^{-5} m

Explanation:

Given:

 Length of cylinder is, L = 0.27 m

Outer radius of cylinder is, r_out = 1.12×10^{-2} m

Inner radius of cylinder is, r_in = 3.9×10^{-3} m

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(a)

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Strain = \frac{Stress}{Young's\ modulus}

     \frac{\Delta L}{L_0}   = \frac{\frac{F}{A}}{Y}

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                  = \frac{(60 kg)(9.8 m/s2 )}{(\pi)[( 1.12\times 10^{-2})^2 - (3.9\times 10^{-3} m)^2] (9.4\times 10^9 N/m2 )}[tex]                   [tex]= 1.80\times 10^{-4} m

(b)  Let assume that humerous is compressed by ΔL

       Since,   strain = ΔL/L0

      (1.80 \times 10^{-4} m) = ΔL / 0.29 m

     Thus,

           ΔL = (4.56 \times 10^{-4} m)(0.29 m)

                  = 5.241 \times 10^{-5} m

7 0
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Rays and wave fronts can generally be used to represent light when the light is interacting with objects that are much larger than the wavelength of light, which is about 500 nm. In particular, we'll use rays and wave fronts to analyze how light interacts with mirrors and lenses.

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