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Rashid [163]
3 years ago
12

On April 15, 1999, a South Korean cargo plane crashed due to a confusion over units. The plane was to fly from Shanghai, China,

to Seoul, Korea. After takeoff plane climbed to 900 m. Then the first officer was instructed by the Shanghai tower to climb to 1500 m and maintain that altitude. The captain. after reaching 1450 m. twice asked the first officer at what altitude they should fly. He was twice told incorrectly they were to be at 1500 ft. The captain pushed the control column quickly forward and started a steep descent. The plane could not recover from the dive and crashed. How much above the correct altitude did the captain think they were when he started his rapid descent and lost control? (It turns out that aircraft altitudes are given in feet throughout the world except in Chine, Mongolia, and the former Soviet states where meters are used.)
Physics
1 answer:
sattari [20]3 years ago
6 0

Answer:

993 m or 3257 ft

Explanation:

The captain was told to fly at at 1500 ft altitude. At the rate of 3.28 ft to 1 m, this is

\dfrac{1500}{3.28}=457 metres.

Since he was at 1450 m, he thought he was above the correct altitude by

1450 - 457 = 993 m.

In feet, this is

993 * 3.28 = 3257 ft

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3 years ago
The drawing shows three objects rotating about a vertical axis. The mass of each object is given in terms of m0, and its perpend
photoshop1234 [79]

Answer:

I₁ > I₃ > I₂

Explanation:

Taking the pic shown, we have

m₁ = 10m₀

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4 years ago
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Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.
Shkiper50 [21]

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

a)

We can solve this first part of the problem by applying the mirror equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

f=\frac{R}{2}

For this mirror, R = 20 cm, so its focal length is

f=\frac{20}{2}=+10 cm (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm

b)

We can solve this part by using the magnification equation:

M=-\frac{y'}{y}=\frac{q}{p}

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

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y = 4 cm

Solving for y', we find the size of the image:

y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm

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6 0
3 years ago
An object starts at 4m/s and accelerates to 6m/s in 10 seconds. Find its displacement.
Natalka [10]

Answer:

Explanation:

The 2 equations we need here are, first:

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Solving for acceleration:

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20 = \frac{2}{5}Δx and

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3 years ago
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