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andreev551 [17]
4 years ago
15

The terminal speed of a sky diver is 130 km/h in the spread-eagle position and 326 km/h in the nosedive position. Assuming that

the diver's drag coefficient C does not change from one position to the other, find the ratio of the effective cross-sectional area A in the slower position to that in the faster position.
Physics
1 answer:
SIZIF [17.4K]4 years ago
8 0

Answer:

\frac{A_{slow} }{A_{fast} }=6.2885

Explanation:

Given data

Terminal velocity for spread eagle position vt=130 km/h

Terminal velocity for nosedive position vt=326 km/h

The terminal speed of the diver is given by

v_{t}=\sqrt{\frac{2mg}{CpA} }

Therefore the area is given by

A=\frac{2mg}{Cp(v_{t})^{2}  }\\

Since everything else is constant in the two dives except for the terminal velocity, the ratio between the area in the slow position to the area in the fast position is

\frac{A_{slow} }{A_{fast} }=\frac{326^{2} }{130^{2} }\\\frac{A_{slow} }{A_{fast} }=6.2885

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The correct option that can be deduced for both Object P and Q is Option b) I and II only

To solve this question correctly, we need to understand the concept of density and it relation to mass and volume.

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From the given parameters, we are being told that:

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This implies that Q has a greater density that P. Since Q has a greater density than P, Q will be heavier since it will have greater mass.

However, Q will not be denser than water because if that happens, P will be have a greater density which is untrue in this scenario.

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Based on their colors, which of the following stars is hottest? Which is coolest? Archenar (blue), Betelgeuse (red), Capella (ye
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Answer: The hottest star is Archenar( blue) and the coolest star is Betelgeuse

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Objects emit radiation that depends exclusively on their temperature. At an ambient temperature, the radiation emitted by an object is in the infrared spectrum (we could only see it with a special camera). If we heat it we will see that it first turns red (whose state we call “red hot”) because it is the lowest and least energetic wavelength of all.

If we continue to heat it, the wavelength that it emits to one with more energy will continue to increase and we will see that it turns yellow and then white. This is a signal that is emitting at all frequencies (but mainly in blue).

If we continue to warm a body that is "white hot", it would emit in the ultraviolet spectrum, with what would become ... black! then we would not see it emits light in the visible spectrum (well, we would see a very faint bluish light corresponding to the tail of the distribution of the spectrum it emits, but the peak of that spectrum would be in the ultraviolet).

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This phenomenon is known as Doppler's effect.

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