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Lina20 [59]
3 years ago
5

Many physiological conditions are related to particular proteins in cell membranes. The number of insulin receptors in membranes

is related to the amount of insulin present around those cells. If the presence of increased insulin results in a decrease in receptors and a decrease in insulin results in an increase in insulin receptors, this would be an example of
Physics
1 answer:
Afina-wow [57]3 years ago
8 0

Answer: This is an example of downregulation and upregulation.

Explanation:

Downregulation is a process in which cells decrease the production of one of their components, responding to an external stimulus. Upregulation, on the other hand, is when cells increase the production of one of their components in response to an external stimulus.

In this case, the decrease in insulin receptors would make the cell less sensitive to the hormone.

If there's a lot of insulin around those cells, the cell would have to decrease its sensitivity, otherwise, it would metabolise more glucose than the body needs. The contrary would happen if there was too little insulin around those cells, they would have to become more sensitive to it by increasing the number of receptors.

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A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i
Arturiano [62]

Answer:

q / q_{1} = 2, assuming that q_{1} and (q - q_{1}) are point charges.

Explanation:

Let k denote the coulomb constant. Let r denote the distance between the two point charges. In this question, neither k and r depend on the value of q_{1}.

By Coulomb's Law, the magnitude of electrostatic force between q_{1} and (q - q_{1}) would be:

\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}.

Find the first and second derivative of F with respect to q_{1}. (Note that 0 < q_{1} < q.)

First derivative:

\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}.

Second derivative:

\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}.

The value of the coulomb constant k is greater than 0. Thus, the value of the second derivative of F with respect to q_{1} would be negative for all real r. F\! would be convex over all q_{1}.

By the convexity of \! F with respect to \! q_{1} \!, there would be a unique q_{1} that globally maximizes F. The first derivative of F\! with respect to q_{1}\! should be 0 for that particular \! q_{1}. In other words:

\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0<em>.</em>

2\, q_{1} = q.

q_{1} = q / 2.

In other words, the force between the two point charges would be maximized when the charge is evenly split:

\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}.

3 0
2 years ago
A 60cm long string of an ordinary guitar is tuned to produce the note a4 (frequency 440hz) when vibrating in its fundamental mod
pishuonlain [190]

Answer:

0.78 m

Explanation:

The relationship between wavelength and frequency of a wave is given by

v=f \lambda

where

v is the speed of the wave

f is the frequency

\lambda is the wavelength

For the sound wave in this problem, we have

f=440 Hz is the frequency

v = 344 m/s is the speed of sound in air

Substituting into the equation and re-arranging it, we find the wavelength:

\lambda=\frac{v}{f}=\frac{344 m/s}{440 Hz}=0.78 m

8 0
3 years ago
Read 2 more answers
Which change in an object would increase the force needed to move the object​
creativ13 [48]

Answer:

force is the answer because force is pushing the item

6 0
3 years ago
What could easily be converted into heat, light, or motion? A) Steel B) Iron C) Electricity D) Energy
julia-pushkina [17]
The answer is C.energy because it can make light and heat
5 0
3 years ago
Write the formula of mechanical advantage​
Zanzabum

Answer:

the formula of mechanical advantage is

MA = load / effort

VR = effort distance / load distance

hope it is helpful to you

7 0
2 years ago
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