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4 years ago

10

How can a protein outside the cell cause events to happen inside the cell?

1 answer:

jeka57 [31]4 years ago

8 0

Answer:

Explained

Explanation:

The proteins(sometimes are called celluar ligands) outside the cell can bind to a receptor protein( generally found on the cell membranes) on the cell surface, causing it( receptor protein) to change the shape and sending a signal inside the cell.

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Is anyone good with 9-10th grade physics?? Please pm me I have a few questions.

Tcecarenko [31]

Let's give it a whirl .....

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3 years ago

Savannah drove 716 miles on Monday. She drove another 572 miles on Tuesday. About how many miles did she drive

Korolek [52]

Explanation:

We have,

Savannah drove 716 miles on Monday and She drove another 572 miles on Tuesday.

It is required to find the total distance covered by her altogether. It is equal to the sum of distance covered on Monday and distance covered on Tuesday. So,

d = 716 miles + 572 miles

d = 1288 miles

So, she drive 1288 miles altogether.

5 0

4 years ago

If the ball leaves the projectile launcher at a speed of 2.2 m/s at an angle of 30ᴼ, and the projectile launcher is on a table a

IgorC [24]

Answer: 1.12 m

Explanation:

This situation is related to parabolic motion, hence we can use the following equations:

$y=y_{o}+V_{o}sin \theta t-\frac{g}{2}t^{2}$ (1)

$x=V_{o} cos \theta t$ (2)

Where:

$y=0 m$ is the ball final height (when it hits the ground)

$y_{o}=1.1 m$ is the ball initial height

$V_{o}=2.2 m/s$ is the initial velocity

$\theta=30\°$ is the angle at which the ball was launched

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the horizontal distance the ball travels

Rewriting (1) with the given values:

$0 m=1.1 m+(2.2 m/s)(cos 30\°)t-\frac{9.8 m/s^{2}}{2}t^{2}$ (3)

Multiplying all the eqquation by -1 and rearranging:

$4.9 m/s^{2} t^{2}-1.1 m/s t-1.1 m=0$ (4)

So, since we have a quadratic equation here (in the form of $0=at^{2}+bt+c$ , we will use the quadratic formula to find $t$ :

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ (5)

Where $a=4.9$ , $b=-1.1$ , $c=-1.1$

Substituting the known values and choosing the positive result of the equation, we have:

$t=\frac{-(-1.1)\pm\sqrt{(-1.1)^{2}-4(4.9)(-1.1)}}{2(4.9)}$

$t=0.59 s$ (6)

Now, substituting (6) in (2):

$x=(2.2 m/s)(cos 30\°)(0.59 s)$ (7)

$x=1.12 m$ (8) This is the horizontal distance at which the ball hits the ground.

3 0

4 years ago

I need an answer to question number 4 ASAP

Nat2105 [25]

Answer:

Screwing

Explanation:

3 0

4 years ago

Encoding information occurs throughout?

alexandr402 [8]

Encoding information occurs throughout forming memory and is the first step of the memory process.

8 0

4 years ago