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3 years ago

What changes would occur at a molecular level if a liquid is placed in cool conditions?

Physics

2 answers:

Hatshy [7]3 years ago

8 0

The molecules will slow down.

Sergeu [11.5K]3 years ago

5 0

C. The motion of the molecules would decrease.

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What is a good example of netwon 3rd law

satela [25.4K]

A fish pushes water backwards in order to move forward is a good example of Newton's 3rd Law.

8 0

3 years ago

What are the divisions within a shell called

blsea [12.9K]

The divisions within an atom's shell are called subshells. This means that each shell consists of several subshells that are made up of orbitals. Each orbital consists of 1 or 2 electrons. The outermost shell of an atom is what we call the valence electrons, and they are what participate in chemical bonding.

5 0

3 years ago

A circular loop of radius r carries a current i. at what distance along the axis of the loop is the magnetic field one-half its

lana [24]

At r = 0.766 R the magnetic field intensity will be half of its value at the center of the current carrying loop.

We have a circular loop of radius ' r ' carrying current ' i '.

We have to find at what distance along the axis of the loop is the magnetic field one-half its value at the center of the loop.

<h3>What is the formula to calculate the Magnetic field intensity due to a current carrying circular loop at a point on its axis?</h3>

The formula to calculate the magnetic field intensity due to a current carrying ( i ) circular loop of radius ' R ' at a distance ' x ' on its axis is given by -

$B(x) = \frac{\mu_{o} iR^{2} }{2(x^{2} +R^{2})^{\frac{3}{2} } }$

Now, for magnetic field intensity at the center of the loop can calculated by putting x = 0 in the above equation. On solving, we get -

$B(0) = \frac{\mu_{o} i}{2R}$

Let us assume that the distance at which the magnetic field intensity is one-half its value at the center of the loop be ' r '. Then -

$\frac{\mu_{o} iR^{2} }{2(r^{2} +R^{2})^{\frac{3}{2} } } = \frac{1}{2} \frac{\mu_{o}i }{2R}$

$2R^{3} = (r^{2} +R^{2} )^{\frac{3}{2} }$

$4R^{6} = (r^{2} +R^{2} )^{3}$

$r^{2} =0.587R^{2}$

r = 0.766R

Hence, at r = 0.766 R - the magnetic field intensity will be half of its value at the center of the current carrying loop.

To solve more questions on magnetic field intensity, visit the link below-

brainly.com/question/15553675

#SPJ4

6 0

2 years ago

The electrical loads in _____ circuits each have the same voltage drop, with equals the total applied voltage of the circuit.

Vesna [10]

Answer:

The electrical loads in parallel circuits each have the same voltage drop, with equals the total applied voltage of the circuit.

Explanation:

I did some research and the voltage drop across any branch of a parallel circuit is the same as the applied voltage.

3 0

2 years ago

A certain field line diagram illustrates the electric field due to three particles that carry charges 5.0 μC, -3.0 μC, and -2.0

4vir4ik [10]

Answer:

Explanation:

Number of lines emanate from + 5 micro coulomb is 15 .

They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.

the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.

So the lines terminating at - 3 micro coulomb

= $\frac{3}{5}\times 15 = 9$

So the lines terminating at - 2 micro coulomb

= $\frac{2}{5}\times 15 = 6$

So, the number of filed lines terminates at - 2 micro Coulomb are 6.

4 0

3 years ago