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1 year ago

Suppose you need to design a parachute system to help a remote camera land safely

Physics

1 answer:

IgorLugansk [536]1 year ago

4 0

The correct answer is hang glider.

A hang-glider cannot take off from low ground since it has no power. It needs to be launched from a high location, such a mountain or a hill. The major force acting on a hang-glider is gravity. The weight of the wing and the pilot together is this. The push that keeps the aerofoil flying through the air is produced by the weight. The hang-aerofoil glider's wing's form prevents it from falling to the ground like a stone. It results in lift. An area of low pressure is created by the aerofoil's acceleration of the air passing over the top of the wing. The air moving beneath the wing is compressed as the wing moves forward and downward. After then, the aerofoil is lifted up into the region of low pressure.

The air will gradually drop if it is still. A hang-glider descends at a speed of roughly 3.6 km/h (slow walking), or about 1 meter per second. A hang-glider needs to locate air coming up at the same rate as the glider is going down in order to maintain height. A hang-glider can fly along a cliff without losing height, for instance, if there is a light breeze coming straight from the sea, the air is being forced vertically upward by the cliff at 3.6 km/h, and the glider is flying over a vertical coastal cliff. The glider will begin to gain altitude in a stronger breeze.

Some hang-glider pilots equip their craft with tiny motors and propellers. They become microlights as a result and can now take off and climb from flat ground like a regular aircraft.

To learn more about hang-glider refer the link:

brainly.com/question/1365947

#SPJ9

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A plane flying horizontally at a speed of 50 m/s and at an elevation of 160 m drops a package, and 2.0 s later it drops a second

Anarel [89]

Answer:

386 m

Explanation:

Let's call the horizontal distance between the point of launch and the point of landing of the first package $x\text{ m}$ .

2 seconds after landing at $x$ , the plane has travelled a horizontal distance of $2\text{ s}\times 50 \text{ m/s} = 100\text{ m}$ .

At this new point, the second package is launched. Because it is launched under the same conditions as the first, its horizontal distance from its point of launch is also $x\text { m}$ .

To determine $x$ , we determine the time of fall of the package. Being a vertical motion, the drop has an initial velocity, $u$ , of 0 m/s with acceleration, $a = g = 9.8 \text{ m/s}{}^2$ , and a distance, $s$ , of 160 m. We use the equation of motion:

$s = ut+\frac{1}{2}at^2$

$160 = 0\times t + \frac{1}{2}9.8\times t^2$

$t = \sqrt{\dfrac{160}{4.9}}\text { s}=\dfrac{40}{7}\text { s}$

In this time, the package, with a horizontal velocity of 50 m/s, travels the horizontal distance, $x$ , of

$x = 50 \text{ m/s}\times \dfrac{40}{7}\text { s} = \dfrac{2000}{7}\text { m} = 286 \text{ m}$

The distance apart between both packages is $x+100\text{ m} = 286+100\text{ m} = 386\text{ m}$

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When titanium bonds with oxygen, the ionic compound that forms has the chemical formula Ti2O3 and consists of Ti3+ ions and O2-

Lesechka [4]

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4 years ago

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Dmitry_Shevchenko [17]

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Andrew [12]

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