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3 years ago

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A plane traveling horizontally at 120 m/s over flat ground at an elevation of 3610 m must drop an emergency packet on a target

on the ground. The trajectory of the packet is given by xequals 120t, yequals minus4.9tsquaredplus3610, for tgreater than or equals 0, where the origin is the point on the ground directly beneath the plane at the moment of the release. How many horizontal meters before the target should the packet be released in order to hit the target? Set up an equation involving t that could be solved in order to find out how long it takes the packet to reach the ground after being released.

2 answers:

antoniya [11.8K]3 years ago

8 0

Answer:

Explanation:

Horizontal displacement

x = 120 t

Vertical position

y = 3610 - 4.9 t²

y = 0 for the ground

0 = 3610 - 4.9 t²

t = 27.14 s

This is the time it will take to reach the ground .

During this period , horizontal displacement

x = 120 x 27.14 m

= 3256.8 m

So packet should be released 3256.8 m before the target.

Pepsi [2]3 years ago

3 0

Answer:

$x=3257.143\ m$

Explanation:

Given:

horizontal velocity of the plane,

elevation of the plane,

Trajectory of the packet is given by:

$x=120.t$ ...........................(1)

$y=-4.9.t^2+3610$ ...................(2)

<u>According to given at ground level we have y = 0.</u>

Putting this condition in eq. (2)

$0=-4.9.t^2+3610$

$t=27.14\ s$

<u>Now the horizontal distance travelled by the plane in the above time:</u>

(Due to inertia of motion the packet will also travel the same distance horizontally from the point where it is dropped.)

$x=120\times 27.14$

$x=3257.143\ m$ is the horizontal distance from the target on the earth before which the packet must be released in order to reach there.

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Answers:

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b) $5.11(10)^{12} cycles$

c) $2.06(10)^{26} cycles$

d) 46000 s

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<h2>a) Time for one cycle of the radio wave</h2>

We know the maser radiowave has a frequency $f$ of $1,420,405,751.786 cycles/s$

In addition we know there is an inverse relation between frequency and time $T$ :

$f=\frac{1}{T}$ (1)

Isolating $T$ : $T=\frac{1}{f}$ (2)

$T=\frac{1}{1,420,405,751.786 cycles/s}$ (3)

$T=7.04(10)^{-10} s$ (4) This is the time for 1 cycle

<h2>b) Cycles that occur in 1 h</h2>

If $1h=3600s$ and we already know the amount of cycles per second $1,420,405,751.786 cycles/s$ , then:

$1,420,405,751.786 \frac{cycles}{s}(3600s)=5.11(10)^{12} cycles$ This is the number of cycles in an hour

<h2>c) How many cycles would have occurred during the age of the earth, which is estimated to be $4.6(10)^{9} years$ ?</h2>

Firstly, we have to convert this from years to seconds:

$4.6(10)^{9} years \frac{365 days}{1 year} \frac{24 h}{1 day} \frac{3600 s}{1 h}=1.45(10)^{17} s$

Now we have to multiply this value for the frequency of the maser radiowave:

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<h2>d) By how many seconds would a hydrogen maser clock be off after a time interval equal to the age of the earth?</h2>

If we have 1 second out for every 100,000 years, then:

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Leno4ka [110]

(a) 3.58 km 45° south of east

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$d=vt$

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t is the time

The average velocity is:

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While we need to convert the time from minutes to seconds:

$t=169 min \cdot 60 s/min = 10140 s$

Therefore, the magnitude of the displacement is

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And the direction is the same as the velocity, therefore 45° south of east.

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The velocity of the air relative to the ground is

$v_a = 2.00 m/s$

and the direction is exactly opposite to that of Allen, so it is 45° north of west. Allen's velocity relative to the ground is

v = 3.53 m/s

So this must be the resultant of Allen's velocity relative to the air (v') and the air's speed ( $v_a$ ). Since these two vectors are in opposite direction, we have

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Therefore we find v', Allen's velocity relative to the air:

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The direction must be measured relative to the air's reference frame. In this reference frame, Allen is moving exactly backward, so his direction will be 90° south of east.

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Answer:

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