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4 years ago

6

I'm not really good on this type of stuff :(

1 answer:

navik [9.2K]4 years ago

5 0

Okay let me walk you through this, a resistor resists right? its in the name. so you can guess that it would be 4th one down, because if you resist you don't stop, you just slow down. So you should definitely know current is energy, and since electrons is energy you know that current is the second one down. You already have two. Heres a good test taking strategy, the first one says "materials that do conduct electricity just not as well as conductors" so you know that something that conducts electricity well is a conductor, the third one down. now you have 3/5 and you haven't used any real knowledge. So now you are stuck between something that blocks electricity and something that is okay with it. Since you know a conductor does good, wouldnt a semi-conductor be the first one? since it is semi or partially a conductor. Then you are left with insulators to block them out. I hope I left a decent enough explanation. :)

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larisa [96]

Answer:

Part a)

$x = 15.76 m$

Part b)

$y = 7.94 m$

Part c)

$x = 26.16 m$

Part d)

$y = 7.49 m$

Part e)

$x = 83.23 m$

Part f)

$y = -75.6 m$

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

$v_x = 19.9 cos39.9$

$v_x = 15.3 m/s$

similarly we have

$v_y = 19.9 sin39.9$

$v_y = 12.76 m/s$

Part a)

Horizontal displacement in 1.03 s

$x = v_x t$

$x = (15.3)(1.03)$

$x = 15.76 m$

Part b)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.03) - 4.9(1.03)^2$

$y = 7.94 m$

Part c)

Horizontal displacement in 1.71 s

$x = v_x t$

$x = (15.3)(1.71)$

$x = 26.16 m$

Part d)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.71) - 4.9(1.71)^2$

$y = 7.49 m$

Part e)

Horizontal displacement in 5.44 s

$x = v_x t$

$x = (15.3)(5.44)$

$x = 83.23 m$

Part f)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(5.44) - 4.9(5.44)^2$

$y = -75.6 m$

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Explanation:

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Answer:

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