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Novay_Z [31]
3 years ago
9

What is Intracellular communication

Physics
1 answer:
Komok [63]3 years ago
8 0

Answer:

Intercellular communication refers to the communication between cells. Membrane vesicle trafficking has an important role in intercellular communications in humans and animals, e.g., in synaptic transmission, hormone secretion via vesicular exocytosis.

You might be interested in
A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.8 1010 m (inside the orbit
creativ13 [48]

Answer:

Explanation:

From the given information:

Distance d_i = 4.8 \times 10^{10} \ m

Speed of the comet V_i = 9.1 \times 10^{4} \ m/s

At distance d_2 = 6 \times 10^{12} \ m

where;

mass of the sun = 1.98 \times 10^{30}

G = 6.67 \times 10^{-11}

To find the speed V_f:

Using the formula:

E_f = E_i + W \\ \\  where; \  \  W = 0  \ \  \text{since work done by surrounding is zero (0)}

E_f = E_i + 0 \\ \\  K_f + U_f = K_i + U_i  \\ \\ = \dfrac{1}{2}mV_f^2 +  \dfrac{-GMm}{d^2} =  \dfrac{1}{2}mV_i^2+ \dfrac{-GMm}{d_i} \\ \\ V_f = \sqrt{V_i^2 + 2 GM \Big [  \dfrac{1}{d_2}- \dfrac{1}{d_i}\Big ]}

V_f = \sqrt{(9.1 \times 10^{4})^2 + 2 (6.67\times 10^{-11}) *(1.98 * 10^{30} ) \Big [  \dfrac{1}{6*10^{12}}- \dfrac{1}{4.8*10^{10}}\Big ]}

\mathbf{V_f =53.125 \times 10^4 \ m/s}

3 0
3 years ago
a 100g ice cube at 0 degrees celsius is placed in 650 grams of water at 25 degrees celsius. When the mixture reaches equillibriu
Artyom0805 [142]

Answer:

The latent heat of fusion of water is 334.88 Joules per gram of water.

Explanation:

Let the latent heat of ice be 'x' J/g

1) Thus heat absorbed by 100 gram of ice to get converted into water equals

Q_1=100\times x

2) heat energy required to raise the temperature of water from 0 to 25 degree Celsius equals

Q_2=100\times 4.186\times 11=4604.6Joules

Thus total energy needed equals Q_1+Q_2=100x+4604.6

3) Heat energy released by the decrease in the temperature of water from 25 to 11 degree Celsius is

Q_3=650\times 4.186\times (25-11)\\\\Q_{3}=38092.6Joules

Now by conservation of energy we have

Q_1+Q_2=Q_3\\\\100x+4604.6=38092.6\\\\\therefore x=\frac{38092.6-4604.6}{100}=334.88J/g

6 0
3 years ago
Consider three scenarios in which a particular box moves downward under the pull of gravity :
luda_lava [24]

Answer:

1. True  WA > WB > WC

Explanation:

In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.

A) Work is the product of force by distance and the cosine of the angle between them

    WA = W h cos 0

   WA = mg h

B) On a ramp without rubbing

     Sin30 = h / L

     L = h / sin 30

     WB = F d cos θ  

     WB = F L cos 30

     WB = mf (h / sin30) cos 30

     WB = mg h ctan 30

C) Ramp with rubbing

    W sin 30 - fr = ma

   N- Wcos30 = 0

   W sin 30 - μ W cos 30 = ma

    F = W (sin30 - μ cos30)

   WC = mg (sin30 - μ cos30) h / sin30

   Wc = mg (1 - μ ctan30) h

When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum

Let's review the claims

1. True The work of gravity is the greatest and the work where there is friction is the least

2 False. The job where there is friction is the least

3 False work with rubbing is the least

4 False work with rubbing is the least

5 0
3 years ago
A box of mass 26 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0
Kazeer [188]

Answer:

\Delta K = 52J

Explanation:

The change in kinetic energy will be simply the difference between the final and initial kinetic energies: \Delta K=K_f-K_i

We know that the formula for the kinetic energy for an object is:

K=\frac{mv^2}{2}

where <em>m </em>is the mass of the object and <em>v</em> its velocity.

For our case then we have:

\Delta K = K_f-K_i=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=\frac{m(v_f^2-v_i^2)}{2}

Which for our values is:

\Delta K = \frac{m(v_f^2-v_i^2)}{2} = \frac{(26Kg)((2m/s)^2-(0m/s)^2)}{2} = 52J

3 0
3 years ago
A ball is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s. How far will the ball go before hit
tia_tia [17]

The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m

s = ut + 1 / 2 at²

s = Distance

u = Initial velocity

t = Time

a = Acceleration

Vertically,

s = 15.4 m

u = 0

a = 9.8 m / s²

15.4 = 0 + ( 1 / 2 * 9.8 * t² )

t² = 3.14

t = 1.77 s

Horizontally,

u = 3.01 m / s

a = 0 ( Since there is no external force )

s = ( 3.01 * 1.77 ) + 0

s = 5.34 m

Therefore, the distance travelled by the ball before hitting the ground is 5.34 m

To know more about distance travelled

brainly.com/question/12696792

#SPJ1

7 0
1 year ago
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