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3 years ago

One of the major categories of receptors in the plasma membrane reacts by forming dimers is

Physics

2 answers:

serg [7]3 years ago

6 0

Answer:

Receptor tyrosine kinases.

Explanation:

Receptor Tyrosine Kinases are cell surface receptors that are attracted to or have an high affinity for hormones, certain growth factors and cytokines that are present in the body.

Receptor Tyrosine Kinases are also known transmembrane receptors. They are activated by signalling mechanisms or pathways.

The particular molecule ( i.e. hormone, growth factors or cytokines) that requires the used of a Receptor Tyrosine Kinases sends a signal which then causes the Receptor Tyrosine Kinases to bind to the surface of the molecules causing a series of reactions to take place in the cell.

Receptor Tyrosine Kinases are very essential and important for:

a. Cell growth

b. Cellular differentiation

c. Cellular metabolism

d. Cellular survival

e. The multiplication or increase in the number of cells also known as proliferation

Receptor Tyrosine Kinases reacts by forming dimers.

ludmilkaskok [199]3 years ago

5 0

Answer:

Receptor tyrosine kinases (RTKs)

Explanation:

Receptor tyrosine kinases is one of the major categories of receptors in the plasma membrane, it reacts by forming dimmers, phosphate groups are also added, and lastly the relay proteins will be activated.

Receptor tyrosine kinases helps in the following cellular processes;

- Motility

- Growth

- Metabolism

- Differentiation.

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Effciency of a lever is never 100% or more. why?Give reason

Troyanec [42]

Answer:

Ideally, the work output of a lever should match the work input. However, because of resistance, the output power is nearly always be less than the input power. As a result, the efficiency would go below $100\%$ .

Explanation:

In an ideal lever, the size of the input and output are inversely proportional to the distances between these two forces and the fulcrum. Let $D_\text{in}$ and $D_\text{out}$ denote these two distances, and let $F_\text{in}$ and $F_\text{out}$ denote the input and the output forces. If the lever is indeed idea, then:

$F_\text{in} \cdot D_\text{in} = F_\text{out} \cdot D_\text{out}$ .

Rearrange to obtain:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}$

Class two levers are levers where the perpendicular distance between the fulcrum and the input is greater than that between the fulcrum and the output. For this ideal lever, that means $D_\text{in} > D_\text{out}$ , such that $F_\text{in} < F_\text{out}$ .

Despite $F_\text{in} < F_\text{out}$ , the amount of work required will stay the same. Let $s_\text{out}$ denote the required linear displacement for the output force. At a distance of $D_\text{out}$ from the fulcrum, the angular displacement of the output force would be $\displaystyle \frac{s_\text{out}}{D_\text{out}}$ . Let $s_\text{in}$ denote the corresponding linear displacement required for the input force. Similarly, the angular displacement of the input force would be $\displaystyle \frac{s_\text{in}}{D_\text{in}}$ . Because both the input and the output are on the same lever, their angular displacement should be the same:

$\displaystyle \frac{s_\text{in}}{D_\text{in}} =\frac{s_\text{out}}{D_\text{out}}$ .

Rearrange to obtain:

$\displaystyle s_\text{in}=s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}$ .

While increasing $D_\text{in}$ reduce the size of the input force $F_\text{in}$ , doing so would also increase the linear distance of the input force $s_\text{in}$ . In other words, $F_\text{in}$ will have to move across a longer linear distance in order to move $F_\text{out}$ by the same $s_\text{out}$ .

The amount of work required depends on both the size of the force and the distance traveled. Let $W_\text{in}$ and $W_\text{out}$ denote the input and output work. For this ideal lever:

$\begin{aligned}W_\text{in} &= F_\text{in} \cdot s_\text{in} \\ &= \left(F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}\right) \cdot \left(s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}\right) \\ &= F_\text{out} \cdot s_\text{out} = W_\text{out}\end{aligned}$ .

In other words, the work input of the ideal lever is equal to the work output.

The efficiency of a machine can be measured as the percentage of work input that is converted to useful output. For this ideal lever, that ratio would be $100\%$ - not anything higher than that.

On the other hand, non-ideal levers take in more work than they give out. The reason is that because of resistance, $F_\text{in}$ would be larger than ideal:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}} + F(\text{resistance})$ .

As a result, in real (i.e., non-ideal) levers, the work input will exceed the useful work output. The efficiency will go below $100\%$ ,

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3 years ago

Faraday's law states that voltage can be changed by moving the coil out of the magnetic field true or false

NNADVOKAT [17]

As per Faraday's law of induction we know that induced EMF in a conducting closed loop is equal to rate of change in flux in that loop

So here we have

$V = \frac{d\phi}{dt}$

now when we move out a coil from magnetic field then in this case there will be EMF induced in that coil as here magnetic flux is changing with time linked with the coil.

Now this induced voltage will remain constant if coil is moved out uniformly

But it will not remain constant if coil is moved out with non uniform speed

So this statement is not always true

so answer must be

FALSE

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WARRIOR [948]

Explanation:

Principle of Floatation

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