How does the egg drop project apply to newtons 3 laws of motion

1 answer:

8 0

The reason why it relates to the newtons 3 laws of motion because the first law of motion states that every object will stay at rest unless it's moved by an unbalanced force which is your hand. The second one states that <span> the velocity of an object changes when it is subjected to an external force meaning it's used by the equation that is commonly used for which is F=M*A. The way it relates to the second law because you are adding force some way or another, the mass is the egg and the acceleration is the drop of the egg while it free falls. And the last one, for a reaction there is always an equal or opposite reaction and the opposite reaction is the floor because it's going against the egg causing it to crack. If it was with the egg, it would have a soft, smooth landing.</span>

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Find time when boy catches the girl or when they are at their closest separation..

soldier1979 [14.2K]

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\implies{Time=10\:sec\:and\:30\:sec}}} \\\\$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}} \\ \\$

$\green{\underline{\bold{Given :}}} \\ \\ \:\:\:\: \bullet\:\:\tt\red{ Velocity \: of \: boy = 50 \: m/s} \\ \\ \:\:\:\: \bullet\:\: \tt\orange{Velocity \: of \: girl = 30 \: m/s }\\ \\ \:\:\:\: \bullet\:\:\tt\green{ Acceleration \: of \: boy = {1 \: m/s}^{2}} \\ \\ \:\:\:\:\bullet\:\: \tt\blue{Acceleration \: of \:girl= {2\: m/s}^{2} }\\ \\ \:\:\:\: \bullet \:\:\tt\purple{Sepration \: between \: them = 150 \: m }\\ \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \\ \:\:\:\: \bullet\:\: \tt\blue{Time \: taken \: to \: catch \: the \: girl }\\$

<u>According to given question</u> :

$\\ \green{\star} \: \text{Using \: relative \: motion \: method} \\ \\ \green{ \circ} \: \tt Net \: velocity = 50 - 30 = 20 \: m/s \\ \\ \green{ \circ} \: \tt Net \: acceleration = 1 - 2 = - 1\: m /{s}^{2} \\\\ \\ \star\:\bold\red{\underline{\:As \: we \: know \: that\:}} \\\\ \tt\purple{: \implies s = ut + \frac{1}{2} {at}^{2}} \\ \\ \tt\green{: \implies 150 = 20 \times t + \frac{1}{2} \times -1 \times {t}^{2}} \\ \\ \tt\purple{: \implies 300 = 40t - {t}^{2}} \\ \\ \tt\green{: \implies {t}^{2} - 40t + 300 = 0} \\ \\ \tt\purple{: \implies t = \frac{ - ( - 40) \pm\sqrt{ { (- 40)}^{2} - 4 \times 1 \times 300} }{2 \times 1} }\\ \\ \tt\green{: \implies t = \frac{40 \pm \sqrt{1600 - 1200} }{2} }\\ \\ \tt\purple{: \implies t = \frac{40 \pm 20}{2} }\\ \\ \green{\tt: \implies t = 10 \: sec \: and \: 30 \: sec}$

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3 years ago

| 12. Which of the following values is equal to one microfarad

goldfiish [28.3K]

For this case we have that by definition, the word "micro", according to the International System, is a prefix that indicates a factor of $10^ {-6}.$ It is represented with the symbol: μ

Therefore, we have that a micro farad is represented as

$1 * 10^{ -6} F$

Answer:

$1 * 10^{ -6} F$

Option A

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WITCHER [35]

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where

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Substituting into the equation, we find

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$\frac{GMm}{R^2}=\frac{mv^2}{R}$