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3 years ago

Can y’all help me answer part 2

Physics

1 answer:

olganol [36]3 years ago

3 0

Hi!

So look at the scatter plot. The youngest and smallest person is number 7. You can tell, because it's at the bottom left corner. The smallest and youngest person is Cathy! So write Cathy for number 7.

Now the tallest people are Alice and Errol. But I can't really tell which one is taller, can you? I can tell that Alice is older. So if I look at the scatter plot, I can see that the two tallest people are 2 and 5, but the older one is 2. Put Alice for 2, and Errol for 5.

Another old person is Dennis. Look at the scatter plot, the second oldest character is 1. Put Dennis for number 1.

The next one is Brenda. She's taller than Cathy and Gavin. So she's the 3rd tallest person. If you look at the scatter plot, you'll see that that is number 4. So put Brenda for number 4.

Now we just have Freda and Gavin. Gavin is in between Cathy and Brenda's height. Which is number 6.

Freda, who is last but certainly not least, is number 3.

Dennis
Alice
Freda
Brenda
Errol
Gavin
Cathy

Hope this helps! :)

-Peredhel

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3 years ago

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mafiozo [28]

Answer:

U = 30 m/s

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The distance covered in this time is the average velocity times time:

Δx = ½ (v + v₀) t

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3 years ago

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2 years ago

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Alex787 [66]

Answer:

F = 296.7N

Explanation:

The x and y component of the position vector are given by:

$x(t) =rcos\phi, y(t) = rsin\phi\\ \frac{dx}{dt} =\frac{dr}{dt} cos\phi - rsin\phi\frac{d\phi}{dt} , \frac{dy}{dt}=\frac{dr}{dt}sin\phi + rcos\phi\frac{d\phi}{dt}\\ \frac{d^2x}{dt^2} = \frac{d^2r}{dt^2} cos\phi - \frac{dr}{dt} sin\phi\frac{d\phi}{dt} -\frac{dr}{dt} sin\phi\frac{d\phi}{dt} - r(cos\phi\frac{d^2\phi}{dt^2} + sin\phi\frac{d^2\phi}{dt^2})$

$\frac{d^2y}{dt^2} = \frac{d^2r}{dt^2} sin\phi + \frac{dr}{dt} cos\phi\frac{d\phi}{dt}+\frac{dr}{dt}cos\phi\frac{d\phi}{dt}+r(-sin\phi\frac{d^2\phi}{dt^2} +cos\phi\frac{d^2\phi}{dt^2})$

At t = 2s:

$\phi(2) = 1.5t^2-6t= -6\\ \frac{d\phi}{dt}(2) = 3t-6=0\\ \frac{d^2\phi}{dt^2}=3\\r(2)=2t+10=14\\ \frac{dr}{dt}=2\\\frac{d^2r}{dt^2} = 0$

Plugging in:

$\frac{d^2x}{dt^2}=-42(cos(-6) + sin(-6))=-52\frac{m}{s^2} \\\frac{d^2y}{dt^2} = 42(cos(-6)-sin(-6))=28.6\frac{m}{s^2}$

The resulting force F is:

$F = m\sqrt{(\frac{d^2x}{dt^2})^2 + (\frac{d^2y}{dt^2})^2}=296.7N$

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3 years ago