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Feliz [49]
3 years ago
6

Although it shouldn’t have happened, on a dive i fail to watch my spg and run out of air. if my buddy is close by, my best optio

n is to another option is to if i’m in shallow water and the surface is closer than my buddy.a) ascend using my buddy's alternate air source / make a controlled emergency swimming ascent
b) ascend using my buddy's alternate air source / make an buoyant emergency ascent
c) make a normal ascent / ascend using my buddy's alternate air source
d) make a controlled emergency swimming ascent / make a normal ascent
Physics
2 answers:
sukhopar [10]3 years ago
8 0

Complete Question:

Although it shouldn't have happened, on a dive I fail to watch my SPG and run out of air. If my buddy is close by, my best option is to __________. Another option is to _____________, if I'm in shallow water and the surface is closer than my buddy.

NB: The options still remain as written in the question

Answer:

a) ascend using my buddy's alternate air source / make a controlled emergency swimming ascent

Explanation:

Since I have exhausted my own air source, my buddy's alternate air source will be the only saver if he is close by. Controlled emergency swimming accent is used if the swimmer's buddy is far but the swimmer is close to the surface. Buoyant emergency ascent can only be used if the swimmer is deep away from the surface and needs to make it to the surface.

leva [86]3 years ago
5 0

Answer:

B ) Ascend using my buddy alternative air source / make an emergency Ascent

Explanation:

From the description it can be seen his buddy is close by of which he can easily use the alternative air source. Also we can see that he is closer to the water surface than his buddy, of which controlled emergency swimming ascent is highly favourable in this condition.

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Morgarella [4.7K]

Answer:

Explanation:

1 )

Put v2,i =0, in second equation

v2,f= (m2-m1)v2,i + 2m1v1,i/m1+m2

v2,f = 0 + 2m1v1,i/m1+m2

v2,f =  2m1v1,i/m1+m2

In this equation coefficient of v1,i is positive so v2,f and v1,i have the same sign.

2 )

Put m1 < m2  and v2,i =0 in first equation

v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2

v1,f= (m1-m2)v1,i

As m1-m2 is negative , v1f and v1i will have opposite sign.

3 )

Put m1 > m2  and v2,i =0 in first equation

v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2

v1,f= (m1-m2)v1,i

m1 - m2 is positive so v1f and v1i will have same  sign.

4 )

Put m1 = m2 and v2,i =0 in first equation

v1,f= (m1-m2)v1,i

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So glider 1 will stop because v1,f = 0 .

 

 

5 0
3 years ago
What is the first step in the formation of a protostar?
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Star formation begins in relatively small molecular clouds called dense cores.[7] Each dense core is initially in balance between self-gravity, which tends to compress the object, and both gas pressure and magnetic pressure, which tend to inflate it. As the dense core accrues mass from its larger, surrounding cloud, self-gravity begins to overwhelm pressure, and collapse begins. Theoretical modeling of an idealized spherical cloud initially supported only by gas pressure indicates that the collapse process spreads from the inside toward the outside.[8] Spectroscopic observations of dense cores that do not yet contain stars indicate that contraction indeed occurs. So far, however, the predicted outward spread of the collapse region has not been observed.[9]

The gas that collapses toward the center of the dense core first builds up a low-mass protostar, and then a protoplanetary disk orbiting the object. As the collapse continues, an increasing amount of gas impacts the disk rather than the star, a consequence of angular momentum conservation. Exactly how material in the disk spirals inward onto the protostar is not yet understood, despite a great deal of theoretical effort. This problem is illustrative of the larger issue of accretion disk theory, which plays a role in much of astrophysics.

Regardless of the details, the outer surface of a protostar consists at least partially of shocked gas that has fallen from the inner edge of the disk. The surface is thus very different from the relatively quiescent photosphere of a pre-main sequence or main-sequence star. Within its deep interior, the protostar has lower temperature than an ordinary star. At its center, hydrogen is not yet undergoing nuclear fusion. Theory predicts, however, that the hydrogen isotope deuterium is undergoing fusion, creating helium-3. The heat from this fusion reaction tends to inflate the protostar, and thereby helps determine the size of the youngest observed pre-main-sequence stars.[11]

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The actual radiation emanating from a protostar is predicted to be in the infrared and millimeter regimes. Point-like sources of such long-wavelength radiation are commonly seen in regions that are obscured by molecular clouds. It is commonly believed that those conventionally labeled as Class 0 or Class I sources are protostars.[12][13] However, there is still no definitive evidence for this identification.

4 0
3 years ago
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harina [27]

I can not solve the problem if I do not have the mass.

3 0
3 years ago
A
loris [4]

Hi

The answer to this question is B. Reaction

7 0
3 years ago
A sound wave traveling at 340 m/s is generated by a 480 Hz tuning fork.
Shtirlitz [24]

Answer:

Wavelength = 0.7083 meters

Explanation:

Given the following data;

Speed of wave = 340 m/s

Frequency = 480 Hz

To find how long is the sound wave, we would determine its wavelength;

Mathematically, the wavelength of a waveform is given by the formula;

Wavelength = velocity/frequency

Wavelength = 340/480

Wavelength = 0.7083 meters

8 0
3 years ago
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