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g100num [7]
3 years ago
8

What is the second law of thermodynamics? Use this law to describe why your feet got cold, not hot, when you put them into the w

ater.
Physics
2 answers:
inna [77]3 years ago
5 0
The second law of thermodynamics has a principle of heat energy travels from higher temperature to lower temperature. This happens when we put our feet into the water. The natural heat of our body will travel to the water, thus we feel cold as the heat on our body travels to the lower temperature of water. 
statuscvo [17]3 years ago
5 0

Explanation:

The second law of thermodynamics describes the flow of heat. Heat flows from an object at higher temperature to the body at colder temperature. It does not flow in the opposite direction.

When you put your feet into the water. The heat from our feet will get transferred to the water. That's why feet gets cold but not get hot when you put them into the water.

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Which molecules dramatically increases the rate of diffusion of water across cell membranes?
Lorico [155]

Answer:

AQUAPORINS dramatically increases the rate of diffusion of water across cell membranes.

Explanation:

Aquaporins are integral membrane proteins also called water channels that forms pores in the membrane mainly for facilitating diffusion of water molecules in and out of the cell. Aquaporins selectively allow the passage of water molecules while preventing the passage of ions and other solutes through it as they are impermeable to protons, and large solutes. These proteins therefore increases the permeability of the membrane to water molecules and the rate of diffusion of water increases through them. Flow of water molecules in aquaporins follows the direction of osmotic pressure across the membrane. Aquaporins in some organisms have been modified to voltage gated aquaporins which helps them to close the water channels in unfavorable conditions such as drought and flooding. This is to prevent excessive loss of water to the surrounding or intake of water into the cells.

5 0
3 years ago
Two equally charged, 2.807 g spheres are placed with 3.711 cm between their centers. When released, each begins to accelerate at
statuscvo [17]
\begin{gathered} m=\text{ 2.807 g} \\ d=\text{ 3.711cm} \\ a=260.125\text{ m/s}^2 \\ m=\text{ mass of  both the spheres} \\ d=\text{ distance between the centers of sphere.} \\ a=\text{ acceleration of spheres.} \end{gathered}\begin{gathered} force\text{ due to the sphere having charge q, outside its surface is given by } \\ \vec{F}=\frac{1}{4\pi\epsilon_o}\frac{q_1q_2}{r^2}\hat{r} \\ q_1=charge\text{ on the source object.} \\ q_2=charge\text{ of the object in which we are observing the force.} \\ F=\text{ the force on the charged particle outside the sphere} \\ r=\text{ distance of the charged particle from the center of the sphere} \\ \hat{r}\text{= direction of the force acting on the charged particle} \end{gathered}\begin{gathered} from\text{ Newton's second law} \\ F=ma \\ F=\text{ force acting on the particle.} \\ m=\text{ mass of the object.} \\ a=\text{ acceleration of the object.} \end{gathered}\begin{gathered} from\text{ both the equation } \\ ma=\frac{1}{4\pi\epsilon_o}\frac{q_1q_{\frac{2}{}}}{r^2}\hat{r} \\ here\text{ q}_1\text{ and q}_2\text{ are the same, according to the question.} \end{gathered}\begin{gathered} converting\text{ all the values in s.i. unit} \\ m=2.807*10^{-3}kg \\ d=3.711*10^{-2}m \\ according\text{ to the question q}_1=\text{ q}_2 \\ value\text{ of }\frac{1}{4\pi\epsilon_o}=9*10^9\text{ Nm}^2\text{/C}^2 \\ now\text{ put all the values in the above equation } \\ 2.807*10^{-3}kg*260.125\text{ m/s\textasciicircum2}=9*10^9Nm^2\text{/C}^2*\frac{q^2}{3.711*10^{-2}m} \\  \end{gathered}\begin{gathered} by\text{ trasformation} \\ q=\sqrt{\frac{2.807*10^{-3}kg*260.125m/s^2*3.711*^10^{-2}m}{9*10^9Nm^2\text{/C}^2}} \\ by\text{ solving this we get } \\ q=17.3514*10^{-7}C \\ q=1.73514\text{ micro coulombs.} \end{gathered}Hence the correct answer is q= 1.73514 micro coulombs.
5 0
1 year ago
Desperate Please Hurry
dusya [7]

Answer:

?????

Explanation:

????????????????????‽,??????????????????‽????????????????????????????????????

5 0
2 years ago
If a 370 KG missile is flying at a speed of 220 m/s , then what is it’s kinetic energy
lidiya [134]

Answer:

8954000 J

Explanation:

The formula to find kinetic energy is:

<em>Kinetic energy = </em>\frac{1}{2}<em> × mass × (velocity)²</em>

So, Kinetic energy = \frac{1}{2} × 370 × (220)²

Kinetic energy = 8954000 J

4 0
2 years ago
A car is sitting still. It accelerates to a constant speed then it decelerates again to zero speed. While the car is acceleratin
Kobotan [32]

Answer:

in the acceleration process the quantity α and w must increase

the deceleration process the alpha quantity must constant  a direction opposite to the angular velocity

Explanation:

Acceleration and angular velocity are related to linear

           v = w xr

            a = αx r

The bold letters indicate vectors and the cross is a vector product, therefore if

we can see that the relationship between linear and angular variables is direct

therefore in the acceleration process the quantity α and w must increase as well as their linear counterparts

in the deceleration process the alpha quantity must constant as the linear acceleration and must have a direction opposite to the angular velocity

5 0
3 years ago
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