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LuckyWell [14K]

3 years ago

11

A feather of mass 0.001 kg falls from a height of 2 m. Under realistic conditions, it experiences air resistance. Based on what

you know about friction, what can you say about the kinetic energy of the feather as it reaches the ground? Acceleration due to gravity is g = 9.8 m/s2.

1 answer:

guajiro [1.7K]3 years ago

7 0

Explanation:

Two objects that only have the force of gravity acting on them, will fall with the same acceleration <span>g=9.8<span>m<span>s2</span></span>; g=32.2<span><span>ft</span><span>s2</span></span></span> and will therefore hit the ground at the same time.

When you drop a feather, air resistance acts on all the surfaces of the feather. This causes the feather to slow down.

Air resistance depends on two factors: the speed of the object (increased for example by throwing it), and its surface area.

hope this can give u a little bet more to know about how to get ur answer :P

have a gud day and gud luck!

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Formula One racers speed up much more quickly than normal passenger vehicles, and they also can stop in a much shorter distance.

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Answer

given,

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Do man made elements have a greater number than natural elements

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in an automobile crash, a vehicle was stopped at a red light is rear-ended by another vehicle. The vehicles have the same mass.

Masja [62]

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Explanation:

Using conservation of momentum :-

$m1u1 + m2v1 = m1u2 + m2v2$

Where:

m1 = Mass of first vehicle

m2 = Mass of second vehicle

u1 = initial speed of first vehicle

v1 = initial speed of second vehicle

u2 = Final speed of first vehicle

v1 = Final speed of second vehicle

From the received informations:

$m1 = m2$

$v1 = 0$

$v2 = u2 = 4 \frac{m}{s}$

So

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3 years ago

If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 64 ft/sec, its height after t seconds

stepan [7]

Answer:

a) $s_{max} = 96\,ft$ , b) $v(4.449\,s) = -78.368\,\frac{ft}{s}$

Explanation:

a) The maximum height is obtained with the help of the First and Second Derivative Tests:

First Derivative

$v(t) = 64 - 32\cdot t$

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Second Derivative

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The maximum height reached by the ball is:

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b) The time required by the ball to hit the ground is:

$32+64\cdot t - 16\cdot t^{2} = 0$

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$t^{2}-4\cdot t - 2 = 0$

$(t -4.449)\cdot (t+0.449)\approx 0$

Just one root offers a solution that is physically reasonable:

$t = 4.449\,s$

The velocity of the ball when it hits the ground is:

$v(4.449\,s) = 64 - 32\cdot (4.449\,s)$

$v(4.449\,s) = -78.368\,\frac{ft}{s}$

6 0

3 years ago

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