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2 years ago

Why is soap solution a better cleansing agent than ordinary water?

1 answer:

8 0

Because it has ingredients in it that will get more Germs off your skin than ordinary water. So, use soap when you wash your hands. XD

You might be interested in

A small but measurable current of 3.8 × 10-10 A exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers

Karolina [17]

Answer:

a) 4.9*10^-6

b) 5.71*10^-15

Explanation:

Given

current, I = 3.8*10^-10A

Diameter, D = 2.5mm

n = 8.49*10^28

The equation for current density and speed drift is

J = I/A = (ne) Vd

A = πD²/4

A = π*0.0025²/4

A = π*6.25*10^-6/4

A = 4.9*10^-6

Now,

J = I/A

J = 3.8*10^-10/4.9*10^-6

J = 7.76*10^-5

Electron drift speed is

J = (ne) Vd

Vd = J/(ne)

Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)

Vd = 7.76*10^-5/1.3584*10^10

Vd = 5.71*10^-15

Therefore, the current density and speed drift are 4.9*10^-6

And 5.71*10^-15 respectively

3 0

2 years ago

Joe is hiking through the woods when he decides to stop and take in the view. He is particularly interested in three objects: a

Novosadov [1.4K]

Answer:

A) correct answer is C, B) correct answer is b and C) The correct answer is b

Explanation:

In the exercises of geometric optics, the equation of the constructor tells us the location of the image.

1 / f = 1 / p + 1 / q

where f is the focal length of the cornea-crystalline system, p and q are the distances to the object and the image.

In this case, the distance to the image on the retina is constant, about 3 cm. Therefore depending on the distance to the object) p = the focal length must change

1 / q = 1 / f 1 / p

let's apply this expression to our case

A) indicates that the tree is at a medium distance

so that the image is formed on the retina THE SAME AS

correct answer is C

B) The squirrel is at a smaller distance (p ') than the tree (p), therefore if we substitute in the equation above we find that q must decrease. Consequently the image is in front of the retina

The mountain is very far, suppose in infinity, so the image is BEHIND THE RETINA

therefore the correct answer is b

C) The squirrel is very close so the curvature of the lens INCREASES, resulting in a DECREASE in the focal length

The correct answer is b

6 0

3 years ago

Which of the following involve an increase in the entropy of the system under consideration? a. melting of a solid b. evaporatio

Mama L [17]

Explanation:

Since, entropy is the measure of degree of randomness. So, more randomly the molecules of a substance are moving more will be its entropy.

For example, when a solid melts then it means heat is absorbed by it due to which its molecules have gained energy. As a result, they collide with each other and hence, entropy will increase.

Evaporation of a liquid will also cause the liquid to change its state from liquid to gas. This means molecules will go far away from each other leading to an increase in the entropy.

Sublimation is a process of conversion of a solid into gaseous phase without going through liquid phase. So, in this case also entropy will increase due to gain in energy by the molecules of a solid.

In freezing, molecules of a substance come closer to each other and acquire less energy. Hence, entropy decreases.

Mixing is a process of combining two or more substances physically with each other. This leads to increase in entropy of a substance.

In separation molecules are separated from each other leading to a decrease in energy. Hence, entropy will also decrease.

Diffusion is a process in which molecules are able to rapidly move from one place to another. Hence, entropy increases when diffusion takes place.

Thus, we can conclude that melting of a solid, evaporation of a liquid, sublimation, mixing and diffusion involve an increase in the entropy of the system under consideration.

8 0

3 years ago

Motor oil , with a viscosity of 0 . 250 Ns / m2 , is flowing through a tube that has a radius of 5 . 00 mm and is 25 . 0 cm long

Thepotemich [5.8K]

Answer:

1.1775 x 10^-3 m^3 /s

Explanation:

viscosity, η = 0.250 Ns/m^2

radius, r = 5 mm = 5 x 10^-3 m

length, l = 25 cm = 0.25 m

Pressure, P = 300 kPa = 300000 Pa

According to the Poisuellie's formula

Volume flow per unit time is

$V=\frac{\pi \times Pr^{4}}{8\eta l}$

$V=\frac{3.14 \times 300000\times \left ( 5\times 10^{-3} \right )^{4}}{8\times 0.250\times 0.25}$

V = 1.1775 x 10^-3 m^3 /s

Thus, the volume of oil flowing per second is 1.1775 x 10^-3 m^3 /s.

3 0

2 years ago

Please view image attached!

Marat540 [252]

Answer:

you would expect a change in the graph

Explanation:

8 0

2 years ago