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givi [52]
3 years ago
6

A drone traveling horizontally at 110 m/s over flat ground at an elevation of 3000 meters must drop an emergency package on a ta

rget on the ground. The trajectory of the package is given by x = 110 t , y = − 4.9 t 2 + 3000 , t ≥ 0 where the origin is the point on the ground directly beneath the drone at the moment of release. How many horizontal meters before the target should the package be released in order to hit the target? Round to the nearest meter.
Physics
1 answer:
zaharov [31]3 years ago
7 0

Answer:

The horizontal distance of the target should be 2721,4 meters.

Explanation:

First of all we need to find the time that the emergency package hits the ground after the moment of release:

y=0 (because when it hits the ground it is on the level of 0m);

0=-4,9*t^2+3000\\t=24,74

The emergency package hits the ground after 24,74 seconds from release.

Lets assume that package preserves his 110 m/s horizontal speed during the free fall. The targets horizontal distance is:

110*24,74=2721,4

2721,4 meters

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A sound wave was determined to have a frequency of 0.3 Hz, speed of 150 cm/s, and amplitude of 2 cm. Find its wavelength.
fredd [130]

Answer:

5 m

Explanation:

From the question,

v = λf....................... Equation 1

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λ = v/f..................... Equation 2

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2 years ago
During the latter part of your European vacation, you are hanging out at the beach at the gold coast of Spain. As you are laying
Jlenok [28]

Well, I guess you can come close, but you can't tell exactly.

It must be presumed that the seagull was flying through the air
when it "let fly" so to speak, so the jettisoned load of ballast
of which the bird unburdened itself had some initial horizontal
velocity.

That impact velocity of 98.5 m/s is actually the resultant of
the horizontal component ... unchanged since the package
was dispatched ... and the vertical component, which grew
all the way down in accordance with the behavior of gravity.

  98.5 m/s  =  √ [ (horizontal component)² + (vertical component)² ].

The vertical component is easy; that's (9.8 m/s²) x (drop time).
Since we're looking for the altitude of launch, we can use the
formula for 'free-fall distance' as a function of acceleration and
time:

             Height = (1/2) (acceleration) (time²) .

If the impact velocity were comprised solely of its vertical
component, then the solution to the problem would be a
piece-o-cake.

                  Time = (98.5 m/s) / (9.81 m/s²) = 10.04 seconds
whence
                 Height = (1/2) (9.81) (10.04)²

                            =   (4.905 m/s²) x (100.8 sec²)  =  494.43 meters.

As noted, this solution applies only if the gull were hovering with
no horizontal velocity, taking careful aim, and with malice in its
primitive brain, launching a remote attack on the rich American.

If the gull was flying at the time ... a reasonable assumption ... then
some part of the impact velocity was a horizontal component.  That
implies that the vertical component is something less than 98.5 m/s,
and that the attack was launched from an altitude less than 494 m.   

8 0
3 years ago
If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 1030 kg, what is the average distance between the planet a
Levart [38]

Answer:

9.36*10^11 m

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To get R we write the formula by making R the subject of the equation

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R³=(G*M*T²)/4π²,

R=∛{(G*M*T²)/4π²},

Substitute values

R=9.36*10^11 m

5 0
2 years ago
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