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igomit [66]
2 years ago
6

Ian walks 2 km to his best friend’s house, then walks 0.5 km to the library. He then makes a 2.5 km walk home. The entire walk t

ook him 3 hours. What is his average speed in km/h?
SHOW YOUR WORK!
Physics
1 answer:
Sonbull [250]2 years ago
4 0

Average speed = (total distance covered) / (time to cover the distance)

Ian's total distance = (2km + 0.5km + 2.5km) = 5 km

Ian's total time = 3 hours

Average speed = (5km) / (3 hours)

Average speed = 1.66 km/hr

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In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp
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The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

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So, first we need to determine the components of the velocity of the ball, like this:

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V_{Ay}=-5.08 m/s

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

x_{A}=V_{Ax}t

x_{A}=(20.38m/s)(0.317s)

x_{A}=6.46m

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

V_{Bx}=20.88 m/s

V_{By}=-2.195 m/s

y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}

0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}

-4.9t^{2}-2.195t+2.1=0

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}

t= -0.9159s    or   t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

x_{B}=V_{Bx}t

x_{B}=(20.88m/s)(0.468s)

x_{B}=9.77m

So once we got the two distances we can now find the difference between them:

x_{B}-x_{A}=9.77m-6.46m=3.31m

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

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Answer:

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