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2 years ago

motorcycle starting from rest cover 200 metre distance in 6 second calculate final velocity and acceleration of the motorcycle

1 answer:

6 0

<h2>Answer:</h2><h2>V=200m</h2><h2><em><u>time=6s</u></em></h2><h2><em><u>time=6s initial velocity=0</u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6</u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s </u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+a</u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value,</u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value, 200/3=0+ax6</u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value, 200/3=0+ax6 200/3-ax6 </u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value, 200/3=0+ax6 200/3-ax6 200/3/6=a </u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value, 200/3=0+ax6 200/3-ax6 200/3/6=a 200/18-a </u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value, 200/3=0+ax6 200/3-ax6 200/3/6=a 200/18-a a 100/9m/s^2</u></em></h2>

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Anti-reflective coatings on lenses use thin-film interference to eliminate the reflection of a particular color. Suppose a glass

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Answer with Explanation:

We are given that

$n_g=1.55$

$n_f=1.35$

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For minimum thickness m=0

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b.Substitute the values

$t=\frac{521\times 10^{-9}}{4\times 1.35}=96.5nm$

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2 years ago

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3 years ago

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2 years ago

A cannon fires a cannonball of mass 16.0 kg by applying a force of 2750 N along the 1.25 m length of the barrel. (a) How much wo

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To develop this problem it is necessary to apply the concepts related to Work and energy conservation.

By definition we know that the work done by a particle is subject to the force and distance traveled. That is to say,

$W = F*d$

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F= Force

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On the other hand we know that the potential energy of a body is given based on height and weight, that is

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$W_{net} = W+PE$

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$W = F*d\\W = 2750*1.25\\W = 3437.5J$

PART B) Applying the height equation and considering that there is an angle in the distance of 25 degrees and the component we are interested in is the vertical, then

$PE = -mgh*sin25$