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mina [271]
3 years ago
14

A jet plane comes in for a downward dive as shown in the figure below.

Physics
2 answers:
xxTIMURxx [149]3 years ago
8 0

The solution would be like this for this specific problem:

<span>5.5 g = g + v^2/r </span><span>
<span>4.5 g = v^2/r </span>
<span>v^2 = 4.5 g * r </span>
<span>v = sqrt ( 4.5 *9.81m/s^2 * 350 m) </span>
v = 124 m/s</span>

So the pilot will black out for this dive at 124 m/s. I am hoping that these answers have satisfied your query and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

wlad13 [49]3 years ago
8 0

As we know that centripetal acceleration during circular motion is given by the formula

a = \frac{v^2}{R}

as we know that maximum value of acceleration is given as

a = 5.50 g

here we know that

g = 9.81 m/s^2

so here we will have

a = 5.50 (9.81 m/s^2)

a = 54 m/s^2

now we will have from above first equation

54 = \frac{v^2}{R}

54 = \frac{v^2}{350}

v = 137.4 m/s

so maximum possible speed will be 137.4 m/s

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Answer:

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).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

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

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

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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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