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igomit [66]
3 years ago
10

Explain and derive the equation for capillary action in the phenomenon of surface tension​

Physics
1 answer:
Mars2501 [29]3 years ago
8 0

Answer:

Explanation:Capillary action is the ability of a liquid to flow in narrow spaces without the assistance of, ... This article is about the physical phenomenon. ... If the diameter of the tube is sufficiently small, then the combination of surface tension (which is caused by cohesion ... They derived the Young–Laplace equation of capillary action.

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If a series circuit contains a 12-V battery, a 6-ohm resistor, and a 4-ohm resistor, what is the current in the circuit?
Shalnov [3]

In a series circuit the total current is the same throughout resistors and so:

I_{total}=I_1=I_2

The voltage is distributed throughout the resistors and so:

V_{total}=V_1+V_2

and the total resistance can be calculated by adding up the resistors resistance:

R_{total}=R_1+R_2

First thing is to calculate the total resistance and so:

R_{total}=6\Omega + 4\Omega = 10\Omega

And by Omh's law V=IR we have:

V_{total}=I_{total}R_{total}\\\\I_{total}=\frac{V_{total}}{R_{total}}= \frac{12V}{10\Omega} =1.2A

And so the total current of the circuit is 1.2 amps i.e. 1.2 A.


6 0
3 years ago
Read 2 more answers
Particles q1, 92, and q3 are in a straight line.
NNADVOKAT [17]

The net force on q₃ will be 17.51 N. The net force is the algebraic sum of the two forces on the pleading q₃

<h3 /><h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The force,by the charge q₁ on the q₃;

\rm F_{31}} = \frac{Kq_1q_3}{r^2} \\\\ \rm F_{31}}  = \frac{9 \times 10^9 \times -28.1 \times 10^{-6}\times \times -47.9 \times 10^{-6}}{(0.600)^2} \\\\ F_{31}} =33.64 \ N

The force,by the charge q₂ on the q₃;

\rm F_{32}} = \frac{Kq_2q_3}{r^2} \\\\ \rm F_{32}}  = \frac{9 \times 10^9 \times 25.5 \times 10^{-6}\times \times -74.9\times 10^{-6}}{(0.300)^2} \\\\ F_{32}} =-19.09  \ N

The net force is the sum of the two forces;

\rm F_{net}=F_{32}+F_{31}\\\\\ \rm F_{net}=36.6-19.9 \\\\ \rm F_{net}=17.51 \ N

Hence, the net force on q₃ will be 17.51 N.

To learn more about Columb's law, refer to the link;

brainly.com/question/1616890

#SPJ1

8 0
2 years ago
If I weighed 130 pounds what would my mass be in kilograms
lawyer [7]
You would weigh 58.967 kilograms
5 0
3 years ago
Read 2 more answers
A table with mass 23.5 kg sits on the ground what is the normal force on the table?
marin [14]

Answer:

C

Explanation:

There are two forces on the table: weight and normal force.  Newton's second law:

∑F = ma

N - mg = 0

N = mg

N = (23.5 kg) (9.81 m/s²)

N = 230 N

6 0
3 years ago
Read 2 more answers
Find the work done by a 75.0 kg person in climbing a 2.50 m flight of stairs at a constant speed.
Lubov Fominskaja [6]

Answer:

<em>1,839.375 Joules</em>

Explanation:

Work is said to be done is the force applied to an object cause the object to move through a distance.

Workdone = Force * Distance

Workdone = mass * acceleration due to gravity * distance

Given

Mass = 75.0kg

acceleration due to gravity = 9.81m/s²

distance = 2.50m

Substitute the given parameters into the formula:

Workdone = 75.0*9.81*2.50

Workdone = 1,839.375Joules

<em>Hence the workdone is 1,839.375 Joules</em>

6 0
3 years ago
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