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Artyom0805 [142]
2 years ago
12

Describe the movement of particles at the melting point of a substance with temperature and energy.

Physics
1 answer:
Gnesinka [82]2 years ago
7 0

Answer:

As a substance reaches the melting point, the particles begin to move faster, causing the substance to become a liquid.

Explanation:

lmk if you need a different or more detailed answer :)

have a wonderous day <3

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After a great many contacts with the charged ball, how is the charge on the rod arranged (when the charged ball is far away)?
faust18 [17]

Answer: Option (b) is the correct answer.

Explanation:

Since, there is a negative charge present on the ball and a positive charge present on the rod. So, when the negatively charged metal ball will come in contact with the rod then positive charges from rod get conducted towards the metal ball.

Hence, the rod gets neutralized. But towards the metal ball there is a continuous supply of negative charges. Therefore, after the neutralization of positive charge from the rod there will be flow of negative charges from the metal ball towards the rod.

Thus, we can conclude that negative charge spread evenly on both ends.

8 0
3 years ago
Can someone tell me everything they know about mechanical waves and ASAP, please!?!?!?
oksano4ka [1.4K]

Answer:

A mechanical wave is a wave that is an oscillation of matter, and therefore transfers energy through a medium. While waves can move over long distances, the movement of the medium of transmission—the material—is limited. Therefore, the oscillating material does not move far fro

Explanation:

6 0
2 years ago
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A biologist looking through a microscope sees a bacterium at r⃗ 1=2.2i^+3.7j^−1.2k^μm(1μm=10−6m). After 6.2 s , it's at r⃗ 2=4.6
stiv31 [10]

The Average velocity for the bacterium is 0.75 unit/sec.

<u>Explanation:</u>

The given values are in the vector form

Where,

dS = distance covered

dT = time interval

Now, to calculate distance covered, we have

|d S|=\sqrt{d S^{2}}

&

d S=r_{2}-r_{1}

d S=(4.6 i+1.9 k)-(2.2 i+3.7 j - 1.2 k)

d S=(4.6-2.2) i+(0-3.7) j+(1.9+1.2) k

d S=2.4 i-3.7 j+3.1 k

Now, putting these values in the standard formula to evaluate the average velocity, we get;

v_{a v g}=\frac{|\mathrm{d} S|}{d T}

v(a v g)=\frac{|\sqrt{\left\{\left(2.4^{2}\right)+\left(3.7^{2}\right)+\left(3.1^{2}\right)\right\}}|}{7.2}

As dT=7.2 sec

Now,

Solving the equation, we get;

v(a v g)=\frac{5.390732789}{7.2}

\begin{aligned}&v(a v g)=\frac{5.39}{7.2}\\&v(a v g)=0.748611111\\&v(a v g)=0.75 \text { units / sec }\end{aligned}

Hence, the average velocity for the bacterium is 0.75 unit/sec.

3 0
3 years ago
An advantage of light microscopes compared to electron microscopes is that light microscopes _____.
ivolga24 [154]
<span>C is the correct answer. Electron microscopes require a vacuum to work, so living cells cannot be seen because they cannot respire. Light microscopes use a ray of visible light instead of a beam of electrons to magnify something so it can be seen by the naked eye. There are two different types of electron microscope: transmission (TEM) and scanning (SEM).</span>
5 0
2 years ago
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We want to design a cylindrical vacuum capacitor, with a given radius a for the outer cylindrical shell, that will be able to st
anastassius [24]

Solution :

a). Using Gauss's law :

  $E=\frac{Q}{4 \pi \epsilon_0r^2}$  ,    $b    .........(1)

Let $E=E_0,\ r=b$ in equation (1)

Therefore, $Q=4 \pi \epsilon_0b^2E_0$  .............(2)

$V_b-V_a = \int^a_b \vec E. d\vec l$

             $=\int^a_b E \ dx$

            $=\frac{Q}{4 \pi \epsilon_0} \int^a_b \frac{1}{x^2} \ dx$

            $=\frac{Q(a-b)}{4 \pi \epsilon_0 a b}$  ....................(3)

Therefore, $U=\frac{1}{2}Q \Delta V$

                     $=\frac{1}{2}(4 \pi \epsilon_0 b^2 E_0)\left(\frac{Q(a-b)}{4\pi \epsilon_0 a b}\right)$

                     $=\frac{4 \pi \epsilon_0}{2a} \ E^2_0 b^3(a-b)$  .............(4)

Now differentiating the equation (4) w.r.t. 'b', we get

$b=\frac{3}{4}a$  

Thus the radius for the inner cylinder conductor is $b=\frac{3}{4}a$

b). For the energy storage, substitute the radius in (4), we get

$U = 4 \pi\epsilon_0 \frac{27a^3E^2_0}{512}$

This is the amount of energy stored in the conductor.

8 0
3 years ago
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