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GalinKa [24]
3 years ago
15

Why do cells have receptors? Receptors make mRNA. Receptors let the cell know when to let things in and out of the cell. Recepto

rs make spike proteins.
Physics
1 answer:
Pavlova-9 [17]3 years ago
6 0

Answer: Cells have receptors because Receptors let the cell know when to let things in and out of the cell.

Explanation:

Cell receptors also called transmembrane receptors are proteins located on the surface of a cell (extracellularly) or inside the cell which receive signals that alters the functions of the cell. The functions of the cells which can be altered includes the alteration in gene transcription and the cell morphology.

Cell receptors are generally categorizes into the following groups:

--> Internal receptors

--> cell surface receptors

--> ion channel receptors

--> G protein coupled receptors

--> enzyme linked receptors

Interaction of cell membrane receptors with specific ligands that bonds to the receptors causes conformational changes in the receptor protein. This in turn, enzymatically activates the intracellular part of the protein or induces interactions between the receptor and the proteins in the cytoplasm that act as second messengers, thereby relaying the signal from the extracellular part of the receptor to the interior of the cell. This enables the cell to know when to let things in or out of it through the information conveyed.

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Since the Wagon is being pulled down hill is it increasing (C)
Ket [755]
<em>Since the wagon is being pulled down hill with a constant velocity, all the forces of the wagon would be (C) increasing.</em>
<em>You are correct! **</em>
4 0
3 years ago
The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge
lesya [120]

Answer:

Loss, \Delta E=-10.63\ J

Explanation:

Given that,

Mass of particle 1, m_1=m =0.66\ kg

Mass of particle 2, m_2=7.4m =4.884\ kg

Speed of particle 1, v_1=2v_o=2\times 6=12\ m/s

Speed of particle 2, v_2=v_o=6\ m/s

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

m_1v_1+m_2v_2=(m_1+m_2)V

V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}

V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}

V = 6.71 m/s

Initial kinetic energy before the collision,

K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)

K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)

K_i=135.43\ J

Final kinetic energy after the collision,

K_f=\dfrac{1}{2}(m_1+m_2)V^2

K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2

K_f=124.80\ J

Lost in kinetic energy,

\Delta K=K_f-K_i

\Delta K=124.80-135.43

\Delta E=-10.63\ J

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0
3 years ago
Megan wants to see if mice the are fed food A or B grow faster rate. She measures the mass of the mice on day 1 then divides the
Setler79 [48]

yes megan does her control is the amount of food given and what food is given between 2 groups and when they are fed all of this concludes to it being a control group hope i helped and please give me brainliest

5 0
3 years ago
Use the conditions provided in the previous problem. Atmosphere statically stable at the base of the mountain, where pressure =
Ann [662]

Answer:

change in relative vorticity  0.0590

Explanation:

Given data

pressure = 1000 hPa

temperature lapse rate q1 = 3.1◦C  per 50 hPa

pressure = 850 hPa

temperature lapse rate q2= -0.61◦C per 50 hPa

to find out

change in relative vorticity

solution

we will apply here formula that is

N = (g /  potential temperature ) × (potential vertical temperature) × exp^1/2    ............................1

here we know g = 9.8 m/s

and q1 = potential temperature=3.3 degree celsius

potential vertical temperature gradient = 3.1 - 0.61  / 1000 -850

potential vertical temperature gradient = 0.0166 degree celsius/hpa

so

N = 9.8 / 2.75 × 0.0166 × exp^1/2

N = 0.0590

8 0
3 years ago
The upward velocity of a 2540kg rocket is v(t)=At + Bt2. At t=0 a=1.50m/s2. The rocket takes off and one second afterwards v=2.0
Alexeev081 [22]

Answer:

The value of A is 1.5m/s^2 and B is 0.5m/s^³

Explanation:

The mass of the rocket = 2540 kg.

Given velocity, v(t)=At + Bt^2

Given t =0  

a= 1.50 m/s^2

Now, velocity V(t) = A*t + B*t²

If,  V(0) = 0, V(1) = 2

a(t) = dV/dt = A+2B × t  

a(0) = 1.5m/s^²  

1.5m/s^²  =  A + 2B ×  0  

A = 1.5m/s^2

now,

V(1) = 2 = A× 1 + B× 1^²  

1.5× 1 +B× 1 = 2m/s

B = 2-1.5  

B = 0.5m/s^³

Now Check V(t) = A× t + B × t^²

So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² ×  1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s  

Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)

7 0
3 years ago
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