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Lera25 [3.4K]
4 years ago
12

A stone is dropped from the top of a tower. What is its velocity after 3.0 seconds?

Physics
1 answer:
kap26 [50]4 years ago
5 0

For purposes of completing our calculations, we're going to assume that
the experiment takes place on or near the surface of the Earth. 

The acceleration of gravity on Earth is about 9.8 m/s², directed toward the
center of the planet.  That means that the downward speed of a falling object
increases by 9.8 m/s for every second that it falls.

3 seconds after being dropped, a stone is falling at (3 x 9.8) = 29.4 m/s. 

That's the vertical component of its velocity.  The horizontal component is
the same as it was at the instant of the drop, provided there is no horizontal
force on the stone during its fall.

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A fighter plane flying at constant speed 420 m/s and constant altitude 3300 m makes a turn of curvature radius 11000 m. On the g
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Answer:

"Apparent weight during the "plan's turn" is  519.4 N

Explanation:

The "plane’s altitude" is not so important, but the fact that it is constant tells us that the plane moves in a "horizontal plane" and its "normal acceleration" is \mathrm{a}_{\mathrm{n}}=\frac{v^{2}}{R}

Given that,

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a_{n}=\frac{420^{2}}{11000}

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It has a horizontal direction. Furthermore, constant speed implies zero tangential acceleration, hence vector a = vector a N. The "apparent weight" of the pilot adds his "true weight" "m" "vector" "g" and the "inertial force""-m" vector a due to plane’s acceleration, vectorW_{\mathrm{app}}=m(\text { vector } g \text { -vector a })

In magnitude,

| \text { vector } g-\text { vector } a |=\sqrt{\left(g^{2}+a^{2}\right)}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{\left(9.8^{2}+16.03^{2}\right)}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{(96.04+256.96)}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{353}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=18.78 \mathrm{m} / \mathrm{s}^{2}

Because vector “a” is horizontal while vector g is vertical. Consequently, the pilot’s apparent weight is vector

\mathrm{W}_{\mathrm{app}}=(18.78 \mathrm{m} / \mathrm{s}^ 2)(53 \mathrm{kg})=995.77 \mathrm{N}

Which is quite heavier than his/her true weigh of 519.4 N

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Question 4 of 10 (1 point) Jump to Question: Choose the word that best completes this sentence. A personal fall arrest system is
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Answer:

The answer is the First One

Explanation:

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