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3 years ago

Describe how transverse waves can be produced by a rope. Then describe hiw pieces of the rope move as waves pass.

1 answer:

3 0

We did this experiment before, when the rope moves, it represents the waves passing through in from the level of intensity. I hope this is a good answer.

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Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic

Nuetrik [128]

Answer:

(a) A = $3.90 \AA$

(b) $A = 4.50 \AA$

(d) $A = 9.02 \AA$

Solution:

As per the question:

Radius of atom, r = 1.95 $\AA = 1.95\times 10^{- 10} m$

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = $2\times 1.95 = 3.90 \AA$

(b) For body centered cubic lattice:

$A = \frac{4}{\sqrt{3}}r$

$A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA$

$A = 2{\sqrt{2}}r$

$A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA$

(d) For diamond lattice:

$A = 2\times \frac{4}{\sqrt{3}}r$

$A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA$

6 0

3 years ago

Will give correct answer brainliest 5 kg m/s 8kg m/s 80 kg m/s 200 kg m/s

o-na [289]

Answer: Here this will help you..

Explanation:

1 kg-m/s to kilogram-force meter/second = 1 kilogram-force meter/second

5 kg-m/s to kilogram-force meter/second = 5 kilogram-force meter/second

10 kg-m/s to kilogram-force meter/second = 10 kilogram-force meter/second

20 kg-m/s to kilogram-force meter/second = 20 kilogram-force meter/second

30 kg-m/s to kilogram-force meter/second = 30 kilogram-force meter/second

40 kg-m/s to kilogram-force meter/second = 40 kilogram-force meter/second

50 kg-m/s to kilogram-force meter/second = 50 kilogram-force meter/second

75 kg-m/s to kilogram-force meter/second = 75 kilogram-force meter/second

100 kg-m/s to kilogram-force meter/second = 100 kilogram-force meter/second

8 0

2 years ago

En una fundición hay un horno eléctrico con capacidad para fundir totalmente 540 kg de cobre. Si la temperatura inicial del cobr

11Alexandr11 [23.1K]

Answer:

Q = 2.95*10^5 kJ

Explanation:

In order to calculate the energy required to melt the cooper, you first calculate the energy required to reach the boiling temperature. You use the following formula:

$Q_1=mc(T_b-T_1)$ (1)

m: mass of cooper = 540 kg

c: specific heat of cooper = 390 J/kg°C

Tb: boiling temperature of cooper = 1080°C

T1: initial temperature of cooper = 20°C

You replace the values of the parameters in the equation (1):

$Q_1=(540kg)(390\frac{J}{kg.\°C})(1080\°C-20\°C)=2.23*10^8J$

Next, you calculate the energy required to melt the cooper by using the following formula:

$Q_2=mL_f$ (2)

Lf: melting constant of cooper = 134000J/kg

$Q_2=(540kg)(134000\frac{J}{kg})=7.24*10^7J$

Finally, the total amount of energy required to melt the cooper from a temperature of 20°C is the sum of Q1 and Q2:

$Q=Q_1+Q_2=2.23*10^8J+7.24*10^7J=2.95*10^8J=2.95*10^5kJ$

The total energy required is 2.95*10^5 kJ

3 0

3 years ago

How do you think festival dances can help you improve your fitness?

Irina18 [472]

Festival dances can help improve your fitness because your moving and exercising your body without even knowing for hours therefore it helps with weight loss and your health in general.

7 0

3 years ago

In the circuit described by the diagram, which pair of resistors is connected in parallel?

scoundrel [369]

Option 4 ( R2 and R3 ) is the correct answer.

Explanation:

In the below given diagram, we can see a circuit diagram that has four resistors such as R1, R2, R3, and R4.
The opening of the circuit is noted as "a" and the ending is noted as "b".
By observing the above diagram, we can clearly see that R2 and R3 are the pair of resistors that are connected in a parallel manner.
Where all the other resistors such as R1 and R4 are neither connected in parallel nor in series.

Hence we can conclude that Resistor R2 and R3 are the ones that are connected in parallel.

6 0

3 years ago