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4 years ago

The smallest group of particles in a crystal that retains the shape of the crystal is called the

Physics

1 answer:

alexdok [17]4 years ago

3 0

Answer:

Unit Cell

Explanation:

A unit cell is the smallest fundamental particle of a crystal structure that represents the structure of a crystal.

Long-range, three-dimensional repetitiveness of this unit cell forms a crystal. A unit cell is composed of ions or atoms in a particular arrangement forming a well-defined structure.

Crystal have a sharp melting and boiling point and when cut using a tool, they give a sharp edge on the pieces.

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How much work is required to compress a gas from a volume of 5.0 l to a volume of 2.5 l by exerting a constant external pressure

ch4aika [34]

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.
Read more on Brainly.com - brainly.com/question/1581851#readmore

4 0

3 years ago

A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit

likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density $\mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m$

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 120=753.6rad/sec$

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by $P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt$

From the relation we can see that power $P\ \propto\ A^2$

So if amplitude is halved then power will be $\frac{1}{4}$ times

So power will be equal to $\frac{0.2023}{2}=0.0505watt$

4 0

3 years ago

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

3 0

4 years ago

The potential energy of a catapult was completely converted into kinetic energy by releasing a small stone with a mass of 20 gra

DerKrebs [107]

1,000 grams = 1 kilogram
20 grams = 0.02 kilogram

Kinetic energy = (1/2) (mass) x (speed)²

(1/2) (0.02) x (15)² =

(0.01) x (225) = 2.25 joules

4 0

3 years ago

Read 2 more answers

Which method will correctly determine whether the forces on an object are balanced or unbalanced?

MaRussiya [10]

For this case, the first thing you should do is define a reference system.

Once the system is defined, we must follow the following steps:

1) Do the sum of forces in a horizontal direction

2) Do the sum of forces in vertical direction

The forces will be balanced if for each direction the net force is equal to zero.

The forces will be unbalanced if for each direction the net force is nonzero.

Answer:

Add the forces in the horizontal and vertical directions separately.

6 0

3 years ago

Read 2 more answers