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Lynna [10]
2 years ago
6

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

in your chaise lounge soaking up the warm Mediterranean sun, a large glob of seagull poop hits you in the face. Since you got an “A” in Physics you are able to estimate the impact velocity at 98.5 m/s. Neglecting air resistance, calculate how high up the seagull was flying when it pooped
Physics
1 answer:
Jlenok [28]2 years ago
8 0

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.   

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For incident ray C, the angle of refraction is 90°. The refracted ray C has the smallest amount of energy of any refracted ray.
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Answer:

Explanation:

When a ray of light travels into rarer medium from the denser medium and the angle of refraction is 90° so the angle of incidence in the denser medium is called critical angle for that pair of media.

Here, the angle of refraction is 90°, so the angle of incidence C is called the angle of incidence which is equal to the critical angle.

8 0
2 years ago
A woman is applying 300N/m2 of pressure on to door with her hand. Her hand has area of 0.02m2. Work out the force being applied​
never [62]

Answer:

6N

Explanation:

Given parameters:

Pressure applied by the woman  = 300N/m²

Area = 0.02m²

Unknown:

Force applied  = ?

Solution:

Pressure is the force per unit area on a body

        Pressure  = \frac{force}{area}

         Force  = Pressure x area

        Force  = 300 x 0.02  = 6N

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A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart, and each chair can seat
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P = 68.125 kW

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Explanation:

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Suppose an X-ray binary is found in which the visible star is a 12 M⦿ red giant, the orbital period is 3.65 days, and the semima
nataly862011 [7]

Answer:

If the mass of a star is greater than 3 solar masses, it will create a black hole. If its mass is less, it will create a neutron star.

Explanation:

If a star's gravity is high enough, when it condenses on itself, it will form a black hole. Otherwise, it will create a large amount of highly dense matter, such as a neutron star. It can be said that if the mass of a star is greater than 3 solar masses, it will create a black hole. If its mass is less, it will create a neutron star.

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A straight wire carries a current of 3 A which is in the plane of this page, pointed toward the top of the page. A particle of c
Ilia_Sergeevich [38]

Answer:

The magnitude of the magnetic force exerted on the moving charge by the current in the wire is 2.18 x 10^{-8} N

The direction of the magnetic force exerted on the moving charge by the current in the wire is radially inward

Explanation:

given information:

current, I = 3 A

q_{0} = +6.5 x 10^{-6} C

r = 0.05 m

v = 280 m/s

and direction of the magnetic force exerted on the moving charge by the current in the wire, we can use the following formula:

F = qvB sin θ

where

F = magnetic force (N)

q = electric charge (C)

v = velocity (m/s)

θ = the angle between the velocity and magnetic field

to find B we use

B = μ_{0}I/2πr

μ_{0} = 4π x 10^{-7} or 1.26 x 10^{-6} N/A^{2} , thus

B = 4π x 10^{-7} x 3 / 2π(0.05)

  = 1.2  x  10^{-5} T

Now, we can calculate the magnitude force

F = qvB sin θ

θ = 90°, because the speed and magnetic are perpendicular

F = 6.5 x 10^{-6} x 280  x 1.2 x  10^{-5} sin 90°

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Using the hand law, the magnetic direction is radially inward

8 0
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