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4 years ago

What might happen if water molecules did not have a slight negative charge on one end and a slight positive charge on the other?

Physics

1 answer:

Katena32 [7]4 years ago

5 0

Answer:

It would not be possible the cohesion among water molecules by the polar covalent bonding.

Well, to understand this in a better way, let's begin by explaining that water is special due to its properties, which makes this fluid useful for many purposes and for the existence of life.

In this sense, one of the main properties of water is cohesion (molecular cohesion), which is the attraction of molecules to others of the same type. So, water molecule ( $H_{2}O$ ) has 2 hydrogen atoms attached to 1 oxygen atom and can stick to itself through hydrogen bonds.

How is this possible?

By the polar covalent bonding, a process in which electrons are shared unequally between atoms, due to the unequal distribution of electrons between atoms of different elements. In other words: slightly positive and slightly negative charges appear in different parts of the molecule.

Now, it can be said that a water molecule has a negative side (oxygen) and a positive side (hydrogen). This is how the oxygen atom tends to monopolize more electrons and keeps them away from hydrogen. Thanks to this polarity, water molecules can stick together.

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A 1,250 W electric motor is connected to a 220 Vrms, 60 Hz source. The power factor is lagging by 0.65. To correct the pf to 0.9

Black_prince [1.1K]

Answer:

$C = 46.891 \mu F$

Explanation:

Given data:

v = 220 rms

power factor = 0.65

P = 1250 W

New power factor is 0.9 lag

we knwo that

$s = \frac{P}{P.F} < COS^{-1} 0.65$

$S = \frac{1250}{0.65} < 49.45$

s = 1923.09 < 49.65^o

s = [1250 + 1461 j] vA

$P.F new = cos [tan^{-1} \frac{Q_{new}}{P}]$

solving for $Q_{new}$

$Q_{new} = P tan [cos^{-1} P.F new]$

$Q_{new} = 1250 [tan[cos^{-1}0.9]]$

$Q_{new} = 605.40 VARS$

$Q_C = Q - Q_{new}$

$Q_C = 1461 - 605.4 = 855.6 vars$

$Q_C = \frac[v_{rms}^2}{xc} =v_{rms}^2 \omega C$

$C = \frac{Q_C}{ v_{rms}^2 \omega}$

$C = \frac{855.6}{220^2 \times 2\pi \times 60}$

$C = 4..689 \times 10^{-5}$ Faraday

$C = 46.891 \mu F$

6 0

4 years ago

Kepler's 3 laws with formula centerpital<br> forces

madreJ [45]

Answer:

Kepler's Third Law

T = 2 π r 3 G M E . T = 2 π r 3 G M E . For an ellipse, recall that the semi-major axis is one-half the sum of the perihelion and the aphelion. For a circular orbit, the semi-major axis (a) is the same as the radius for the orbit.

6 0

3 years ago

Technician A says that galvanic activity is the result of a poor ground connection. Technician B says that electrolysis can occu

Drupady [299]

Answer:

Only B is correct.

Explanation:

Electrolysis and galvanic activity are two different ways of the passage of electricity. Electrolysis occurs at a spontaneous reaction, and galvanic activity needs an external force to happen.

For example, a cell works by a redox reaction, in which a material reduces (gain electrons), and the other oxides (loses electrons). As higher is the standard reduction potential of a material, as easy is to it to reduces. Thus, if a cell works with the reduction happens with the material with low standard reduction potential, then it works by galvanic activity.

When electronics are connected to the house system, it's necessary to do the ground connection, which will allow the electrons that are dissipated to go to the earth. So, it is done incorrectly, the electrons flow to the equipment, which they natural path. So, the electrolysis can occur with a poor ground connection.

7 0

4 years ago

What is umax,c, the value of the maximum energy stored in the capacitor during one cycle?

TiliK225 [7]

1) 0.266 H

2) 0.040 J

3) $0.308\Omega$

Explanation:

The diagram of the circuit is missing: find it in attachment.

1) What is L, the value of the inductance of the circuit?

For an RLC circuit, the resonant angular frequency is given by:

$\omega=\frac{1}{\sqrt{LC}}$

where

L is the inductance of the circuit

C is the capacitance of the circuit

In this problem, we have:

$\omega=164 rad/s$ is the angular frequency of the generator, at which the circuit is in resonance

$C=140 \mu F = 140\cdot 10^{-6}F$ is the capacitance

Therefore, solving for L, we find the inductance of the circuit:

$L=\frac{1}{\omega^2 C}=\frac{1}{(164)^2(140\cdot 10^{-6})}=0.266 H$

2) What is umax,c, the value of the maximum energy stored in the capacitor during one cycle

The maximum energy stored in a capacitor is given by the equation

$U=\frac{1}{2}CV^2$

where

C is the capacitance of the capacitor

V is the maximum potential difference across the capacitor

Here we have:

$C=140 \mu F = 140\cdot 10^{-6}F$ is the capacitance

$V=24 V$ is the maximum voltage across the capacitor, which is the emf of the generator

Substituting, we find:

$U=\frac{1}{2}(140\cdot 10^{-6})(24)^2=0.040 J$

3) What is R, the value of the resistance of the circuit?

Here we want to find the resistance of the circuit.

The resistance of the circuit can be found by using Ohm's Law:

$V=RI$

where

V is the maximum voltage

R is the resistance

I is the maximum current

Here we have:

V = 24 V is the maximum voltage provided by the generator

I = 0.78 A is the maximum current in the circuit

Solving for R, we find the resistance:

$R=\frac{V}{I}=\frac{0.24}{0.78}=0.308\Omega$

6 0

4 years ago

90cm uniform lever has a load 30N suspended at 15cm from one of it's end. If the fulcrum is at the center of gravity. The force

solniwko [45]

Answer: F = 20 N

Explanation:

I will ASSUME that the fulcrum is at the center of gravity of the lever arm, This means that the lever arm itself creates no moment about the fulcrum because there is no moment arm for that particular force.

To solve, we sum moments about any convenient point to zero (zero because there is no acceleration in the F = ma equation)

The easiest convenient point is the fulcrum

30((90/2) - 15) - F(90/2) = 0

30(30) = F(45)

F = 900/45 = 20 N

3 0

3 years ago