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3 years ago

Function Of The glottis

Physics

2 answers:

Cloud [144]3 years ago

8 0

Answer:

Their function is to produce sound by allowing the free edges of the folds to vibrate against one another and also to act as the laryngeal sphincter when they are closed.

Lunna [17]3 years ago

7 0

Answer:

The glottis is found within the larynx, and the epiglottis is a flap-like structure on top of the trachea. When swallowing, it will press downward to partially close off the trachea. Function of the glottis and epiglottis do to help prevent food and liquid from entering the air passage (trachea)

Can I please get brainliest

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Some believe that the positions of the planets at the time

laila [671]

Answer:

$1.4007\times 10^{-8}\ N$

$1.50075\times 10^{-6}\ N$

$0.000000667\ N$

Explanation:

$m_1$ = Mass of baby = 3 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Distance between objects

Gravitational force of attraction is given by

$F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 70\times 3}{1^2}\\\Rightarrow F=1.4007\times 10^{-8}\ N$

The force between baby and obstetrician is $1.4007\times 10^{-8}\ N$

$F=\dfrac{6.67\times 10^{-11}\times 3\times 2.7\times 10^{27}}{(6\times 10^{11})^2}\\\Rightarrow F=1.50075\times 10^{-6}\ N$

The force between the baby and Jupiter is $1.50075\times 10^{-6}\ N$

$F=\dfrac{6.67\times 10^{-11}\times 3\times 2.7\times 10^{27}}{(9\times 10^{11})^2}\\\Rightarrow F=0.000000667\ N$

The force between the baby and Jupiter is $0.000000667\ N$

7 0

3 years ago

A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s

Paraphin [41]

Answer:

$\omega = 0.016\,\frac{rad}{s}$

Explanation:

The rotation rate of the man is:

$\omega = \frac{v}{R}$

$\omega = \frac{0.80\,\frac{m}{s} }{5\,m}$

$\omega = 0.16\,\frac{rad}{s}$

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

$(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega$

The final angular speed is:

$\omega = 0.016\,\frac{rad}{s}$

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3 years ago

How much force is required to accelerate a mass of 12kg at 3.0 m/s ^ 2

konstantin123 [22]

This is the answer your welcome

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2 years ago

On the incomplete ray diagram for an object in front of a curved mirror, trace the path of Ray 1 as you would do to find the loc

Alisiya [41]

Answer:

See attached the ray diagram with the path of Ray 1, which must pass through the focal point, F.

Explanation:

The curved mirror shown is a converging lens.

To find the location and size of the image formed by a converging lens, you must draw two rays: one ray is from the upper tip of the object (the pen in your figure) that is in front of the mirror, parallel to the horizontal axis until the lens (Ray 1 in your figure) which then bends through the focal point (F in your figure). That is the ray that you must complete in your figure.

The attached figure shows this ray in blue.

The other ray would be from the upper tip of the pen straight through the center of the lens (this is not included in the figure, by instructions of the question).

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3 years ago

A 2 kg ball travellng to the right with a speed of 4 m/s collidees with a 5 kg ball traveling to the left with a speed of 3 m/s.

Savatey [412]

Explanation:

Momentum before collision:

(2 kg) (4 m/s) + (5 kg) (-3 m/s) = -3 kg m/s

No external forces act on the balls, so momentum is conserved. Therefore, momentum after collision is also -3 kg m/s.

5 0

3 years ago

Function Of The glottis​

Function Of The glottis