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SpyIntel [72]
3 years ago
6

Which of the following muscles act(s) to increase the size of the thoracic cavity during inhalation? 1. diaphragm 2. external in

tercostals 3. internal intercostals
Biology
1 answer:
Gemiola [76]3 years ago
3 0

Answer: External intercoastal muscles increase the size of thorasic cavity during inhalation.

Explanation:

Inhalation is the process of breathing in oxygen.

During inhalation, diaphragm contracts and it lead to the expansion of lung volume. When diaphragm contracts, it moves to the abdominal cavity which create a large thorasic cavity . when the external intercoastal muscles contracts, it cause toward and downward movement of the ribs, which cause the rib cage to expand , thereby leading to increase in size of thorasic cavity.

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Pls help ill give brainkliest and 5 stars
rewona [7]

Picture 1 depicts convergent boundary.

Picture 2 depicts divergent boundary.

Picture 3 depicts transform boundary.

<h3>What is a Divergent boundary?</h3>

This occurs when two tectonic plates move away from each other while convergent boundary occurs when the plates collide with each other.

Transform boundary is a fault along a plate boundary where the motion is mostly horizontal.

Read more about Divergent boundary here brainly.com/question/8866854

4 0
3 years ago
An owl has good night vision because its eyes can detect a light intensity as small as 5.0 × 10-13 W/m2. What is the minimum num
Lady bird [3.3K]

Answer:

Thus, the minimum number of photons per second is 77.34

Explanation:

Light intensity, I_{min} =  5\times 10^{-13} W/m^{2}

Pupil has a diameter, d = 8.5 mm

                                      = 8.5 x 10^{-3} m

Radius of the eye, r = 4.25 x 10^{-3} m

∴ Area of the eye, A = \pi .r^{2}

                                 = 3.14\times \left ( 4.25\times 10^{-3} \right )^{2}

                                = 5.6\times 10^{-5} m^{2}

Let P_{min} be the minimum number of photons.

Therefore, P_{min} = I_{min} x A

                                              = 5\times 10^{-13} x 5.6\times 10^{-5}

                                             = 2.8\times 10^{-17} W

Thus the minimum number of photons is given by

N_{min}=P_{min}/E

where E = hc/\lambda

             = \left (6.63\times 10^{-34}\times 3\times 10^{8}  \right )/548\times 10^{-9}

            = 3.62\times 10^{-19} J

Therefore, N_{min} = \frac{2.8\times 10^{-17}}{3.62\times 10^{-19}}

                                              = 77.34 photons per second

Thus, the minimum number of photons per second is 77.34

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Which disease is caused indirectly by pollution? ebola cholera influenza aids
valina [46]
Cholera, its a water borne disease.
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If you didn't have macrophages in your body then,
Dovator [93]
<span>You wouldn't be able to defend against germs</span>
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Which of these sentences is true about the virus?select one of the options below as your answer:
Dima020 [189]
When infected by a virus, a host cell produce many thousands of identical copies of the original virus. so the answer is B 
hope this helps:)
5 0
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