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Irina18 [472]
3 years ago
14

What is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?

Physics
1 answer:
Art [367]3 years ago
3 0

CORRECT ANSWER:

a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.

STEP-BY-STEP EXPLANATION:

The complete question from book is

According to Figure 9.6, what is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?

a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.

b- Signaling molecules that bind to cell-surface receptors lead to cellular responses restricted to the cytoplasm; signaling molecules that bind to intracellular receptors lead to cellular responses restricted to the nucleus.

c- Cell-surface receptors bind to specific signaling molecules; intracellular receptors bind any signaling molecule.

d- Cell-surface receptors typically bind to signaling molecules that are smaller than those bound by intracellular receptors.

e- None of the other answer options is correct.

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A car is traveling at 8 m/s accelerates at 3.1 m/s^2 to reach a final top speed of 56 m/s. How much time will pass before the ca
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Please find attached photograph for your answer.

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The number of sound waves per unit time is called what
Andreyy89

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Frequency

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A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit
Ratling [72]

Answer:

The magnitude of angular acceleration is 232.38\ rad/s^2.

Explanation:

Given that,

Initial angular velocity, \omega_i=3500\ rev/min=366.5\ rad/s

When it switched off, it comes o rest, \omega_f=0

Number of revolution, \theta=46=289.02\ rad

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2  

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6 0
3 years ago
A sprinter starts from rest and runs with constant acceleration to a top speed of 12m/s in 4.0 seconds.
bagirrra123 [75]

Answer:

  a =  3,0 m/s²

Explanation:

En este ejercicio se pide calcular la aceleracion del cuerpo, usemos las ecuaciones de cinematica en una dimensión.

          v= v₀ + a t

como el corredor parte del reposo si velocidad inicial es cero

           v =  at

           a = v/t

calculemos

          a = 12 /4,0

          a =  3,0 m/s²

4 0
3 years ago
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