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Oliga [24]
3 years ago
7

Two arrows are shot vertically upward. the second arrow is shot after the first one, but while the first is still on its way up.

the initial speeds are such that both arrows reach their maximum heights at the same instant, although these heights are different. suppose that the initial speed of the first arrow is 31.8 m/s and that the second arrow is fired 1.82 s after the first. determine the initial speed of the second arrow.
Physics
1 answer:
Neko [114]3 years ago
7 0
<span>Now that you know the time to reach its maximum height, you have enough information to find out the initial velocity of the second arrow. Here's what you know about it: its final velocity is 0 m/s (at the maximum height), its time to reach that is 2.8 seconds, but wait! it was fired 1.05 seconds later, so take off 1.05 seconds so that its time is 1.75 seconds, and of course gravity is still the same at -9.8 m/s^2. Plug those numbers into the kinematic equation (Vf=Vi+a*t, remember?) for 0=Vi+-9.8*1.75 and solve for Vi to get....... 17.15 m/s</span>
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3. A concrete highway is built of slabs 12 m long (20o C). How wide should the expansion cracks between the slabs be (at 20o C)
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Answer:

\Delta x=2.304\times 10^{-2}\ m=2.304\ cm is the minimum gap between the slabs

Explanation:

Given:

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<u>Now form the equation of thermal expansion:</u>

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\Delta l=1.152\times 10^{-2}\ m

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\Delta x=2\times \Delta l

\Delta x=2\times( 1.152\times 10^{-2})

\Delta x=2.304\times 10^{-2}\ m=2.304\ cm is the minimum gap between the slabs

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The detailed calculations are shown below;

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  It is seen that the maximum change in velocity is at the time between 8s to 10s.

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c)The total distance traveled by the particle = Area under the curve

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