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2 years ago

11

A plane drops a rubber raft to the survivors of a shipwreck. The plane is flying at a height of 1960 m and with a speed of 109 m

/s. The raft lands next to the survivors. Ignoring air resistance, how far away from the shipwreck was the plane when the raft was dropped? (g = 9.8 m/s2) (Round your answer to a whole number).

1 answer:

Korolek [52]2 years ago

8 0

In this case the rubber raft has horizontal and vertical motion.

Considering vertical motion first.

We have displacement $s = ut +\frac{1}{2}at^2$ , u = Initial velocity, t = time taken, a = acceleration.

In vertical motion

s = 1960 m, u = 0 m/s, a = 9.81 $m/s^2$

$1960 = 0*t+\frac{1}{2} *9.81*t^2\\ \\ t = 20 seconds$

So raft will take 20 seconds to reach ground.

Now considering horizontal motion of raft

u = 109 m/s, t = 20 s, a = 0 $m/s^2$

So $s = 109*20+\frac{1}{2} *0*20^2 = 2180 m$

So shipwreck was 2180 meter far away from the plane when the raft was dropped.

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Explanation:

From the question we are told that

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So

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Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

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3 years ago

Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi

kozerog [31]

Answer:

a) (0, -33, 12)

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Explanation:

<h2>a) </h2>

We know that the cross product of linearly independent vectors $\vec{A}$ and $\vec{B}$ gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors $\vec{A} = \vec{P}-\vec{Q}$ and $\vec{B} = \vec{R} - \vec{Q}$ are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

$\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)$

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$\vec{A} \times \vec{B} = (A_y B_z - B_y A_z) \ \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( (-1) * 3 - 1 * (-3) ) \ \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( - 3 + 3 ) \ \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}$

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<h2>B)</h2>

We know that $\vec{A}$ and $\vec{B}$ are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

$|\vec{A} \times \vec{B} | = 2 * area_{triangle}$

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$\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}$

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pentagon [3]

Answer:

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Alekssandra [29.7K]

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