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2 years ago

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

aste receptors

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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A small person was running way faster than a bigger person that weighed more collided in a football game. Who would be pushed ba

Semenov [28]

Answer:

smaller one

Explanation:

even though he is moving quicker doesn't mean he will be packing more force in the collision

5 0

3 years ago

tom does not reallt want to give away blue marbles and would like to change the probability that he chooses a blue marble to one

Llana [10]

We don't know how many of ANY color are in the bag right now, so there's no way to calculate an answer.

What Tom has to do is make sure that the number of marbles that are NOT blue is NINE TIMES the number of blue ones in the bag.

4 0

3 years ago

Will mark as brainliest if correct!!!!!!!

lutik1710 [3]

C. A could be ruby (speed of light = 170,000 km/s); B could be diamond (speed of light = 120,000 km/s).

Explanation:

Refraction is a phenomenon that occurs when a light rays crosses the boundary between two different mediums.

When this occurs, the light wave changes speed and also direction, according to Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$ (1)

where

$n_1$ is the index of refraction of the first medium

$n_2$ is the index of refraction of the second medium

$\theta_1$ is the angle of incidence (the angle between the incident ray and the normal to the boundary)

$\theta_2$ is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

The index of refraction is the ratio between the speed of light in a vacuum (c) and the speed of light in the medium (v):

$n=\frac{c}{v}$

Using this definition, we can rewrite eq.(1) as

$\frac{\sin \theta_2}{\sin \theta_1} = \frac{v_2}{v_1}$

Where $v_2$ is the speed of light in the 2nd medium and $v_1$ the speed of light in the 1st medium.

Now let's analyze the situation represented in the figure: we see that as the light ray enters the 2nd medium, it bends towards the normal, this means that the angle of refraction is smaller than the angle of incidence:

$\theta_2 < \theta_1$

This means that

$\frac{\sin \theta_2}{\sin \theta_1}$

And therefore,

$\frac{v_2}{v_1}$

So, the speed of light in the second medium is smaller than the speed of light in the first medium: this occurs only in option C), which is therefore the correct choice.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

6 0

3 years ago

Suppose that the sound level of a conversation is initially at an angry 70 db and then drops to a soothing 50 db. assuming that

stepan [7]

Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2

Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2

Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2

Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2

Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]

Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm

Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm

4 0

3 years ago

As the concentration of a solute in a solution increases, the freezing point of the solution ________ and the vapor pressure of

kykrilka [37]

Answer:

As the concentration of a solute in a solution increases, the freezing point of the solution <u><em>decrease </em></u>and the vapor pressure of the solution <em><u>decrease </u></em>.

Explanation:

Depression in freezing point :

$\Delta T_f=K_f\times m$

where,

$\Delta T_f$ =depression in freezing point =

$K_f$ = freezing point constant

m = molality ( moles per kg of solvent) of the solution

As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:

If molality of the solution in high the depression in freezing point of the solution will be more.
If molality of the solution in low the depression in freezing point of teh solution will be lower .

Relative lowering in vapor pressure of the solution is given by :

$\frac{p_o-p_s}{p_o}=\chi_{solute}$

$p_o$ = Vapor pressure of pure solvent

$p_s$ = Vapor pressure of solution

$\chi_{solute}$ = Mole fraction of solute

$p_s\propto \frac{1}{\chi_{solute}}$

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.

Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.

8 0

2 years ago