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raketka [301]
2 years ago
12

Which receptors are responsible for the production of saliva (A) auditory receptors (B) optic receptors (C) skin receptors (D) t

aste receptors
Physics
1 answer:
Fudgin [204]2 years ago
8 0

Answer:

\large \boxed{\sf D. \ taste \ receptors}

Explanation:

When activated, the receptor most likely prompting the production of saliva is the taste receptor. When food enters the mouth, the salivary glands produce the saliva upon the sensation of taste.

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Which is true of electricity generated both from coal and from nuclear reactions?
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They both release greenhouse gases, I think
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When a user wants to participate in a PKI, what component does he or she need to obtain, and how does that happen?
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Sb +<br> Cl2 →<br> SbCl3<br> Balance chemical equation
melamori03 [73]

Answer:

Cl2 + Sb → SbCl3

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Are predators or their prey more likely to be successful?
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An oscillator consists of a block attached to a spring (k = 500 N/m). At some time t, the position (measured from the system's e
Alex_Xolod [135]

Answer:

a) \omega = 10.407\,\frac{rad}{s}, b) m = 4.617\,kg, c) A = 1.355\,m

Explanation:

a) The system have a simple armonic motion, whose position function is:

x(t) = A\cdot \cos (\omega\cdot t + \phi)

The velocity function is determined by deriving the position function in terms of time:

v(t) = -\omega \cdot A \cdot \sin(\omega\cdot t + \phi)

The acceleration function is found by deriving again:

a(t) = -\omega^{2} \cdot A \cdot \cos (\omega\cdot t + \phi)

Let assume that t = 0\,s. The following nonlinear system is built:

A\cdot \cos \phi = 0.660\,m

-\omega \cdot A \cdot \sin \phi = -12.3\,\frac{m}{s}

-\omega^{2}\cdot A \cdot \sin \phi = -128\,\frac{m}{s^{2}}

System can be reduced by divinding the second and third expressions by the first expression:

\omega \cdot \tan \phi = 18.636\,\frac{1}{s}

\omega^{2}\cdot \tan \phi = 193.94\,\frac{1}{s^{2}}

Now, the last expression is divided by the first one:

\omega = 10.407\,\frac{rad}{s}

b) The mass of the block is:

m = \frac{k}{\omega^{2}}

m = \frac{500\,\frac{N}{m} }{(10.407\,\frac{rad}{s})^{2} }

m = 4.617\,kg

c) The phase angle is:

\phi = \tan^{-1} \left(\frac{18.636\,\frac{1}{s} }{\omega}  \right)

\phi \approx 0.338\pi

The amplitude is:

A = \frac{0.660\,m}{\cos 0.338\pi}

A = 1.355\,m

8 0
3 years ago
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