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krok68 [10]
3 years ago
15

Why do you need to breath faster at the top of a mountain

Physics
1 answer:
Dovator [93]3 years ago
6 0
Because of the higher elevation. As much as we go away from earth by increasing altitude, there will occur lacoccuren. due to this, we need to breathe fast to get suffient oxygen by time at top of the mountain. We breathe faster at the top of a tall mountain due to a natural reflex. As we go higher up a mountain, the density of air and oxygen decreases, i.e. the amount of oxygen in the air decreases along with other gases. Therefore, to maintain the regular supply of oxygen, the body breathes faster. If we go too high without oxygen tanks, we could fall unconscious and die within 8 minutes due to oxygen deprivation.

Hope this helps! let me know if you need anything else!!

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2)It is known that the connecting rodS exerts on the crankBCa 2.5-kN force directed down andto the left along the centerline ofA
11111nata11111 [884]

Answer:

M_c = 100.8 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two cases shown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

                        M_c = F_a*( 42 / 150 ) *144

                        M_c = 2.5*( 42 / 150 ) *144

                        M_c = 100.8 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

5 0
3 years ago
Which describes the characteristics of a liquid?
Talja [164]
What happens when you pour water in a glass? It takes the shape of the glass. This means that water can't have a fixed shape or volume

5 0
3 years ago
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3 0
3 years ago
An airplane traveling at one third the speed of sound (i.e., 114 m/s) emits a sound of frequency 3.72 kHz. At what frequency doe
nlexa [21]

Answer:

f'=5.58kHz

Explanation:

This is an example of the Doppler effect, the formula is:

f'=\frac{(v+v_o)}{(v+v_s)}f

Where f is the actual frequency, f' is the observed frequency, v is the velocity of the sound waves, v_o the velocity of the observer (which is negative if the observer is moving away from the source)  and v_s the velocity of the source  (which is negative if is moving towards the observer). For this problem:

f=3.72kHz\\v=342m/s\\v_o=0m/s\\v_s=-114m/s

f'=\frac{(342+0)}{(342-114)}3.72\times10^3\\f'=\frac{342}{228}3.72\times10^3\\f'=(1.5)3.72\times10^3\\f'=5580Hz=5.58kHz

5 0
3 years ago
The diagram shows a wave traveling through a medium.
m_a_m_a [10]
The answer is b. The crest.

Hope this helps!
6 0
3 years ago
Read 2 more answers
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