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Nikolay [14]
3 years ago
13

Indica qué es una propiedad específica de la materia. Además explica por qué son útiles las propiedades específicas de la materi

a frente a las generales.
Physics
1 answer:
Katyanochek1 [597]3 years ago
7 0

Answer:

Check Explanation

Comprobar explicación

Explanation:

English Translation

Indicate what a specific property of matter is. Also explain why the specific properties of matter are useful compared to the general ones.

Solution

The specific properties of matter are properties that describes the intensive properties of the system. They are properties that do not depend on or change with the extent or size of the system. They are usually obtained by dividing the generalised properties or extensive properties by the extent or size of matter to make them independent of size/extent/Mass.

Examples of specific properties include specific heat capacity, specific volume etc. They usually have units of general units/Mass units.

The specific properties of matter are more important than the general ones because

- They help in general comparisons of the properties of different materials. They are used to rank, classify and compare properties of different materials.

- They are used in reference table/data to easily record easily accessible properties of matter. It helps to record standards that are general and independent of sizes/extents/Mass, thereby keeping the reference table/data/chart precise and concise.

- They provide us with values that are easy to memorize and remember, unlike trying to cram the different properties of different masses/sizes of matter.

In Spanish/En español

Las propiedades específicas de la materia son propiedades que describen las propiedades intensivas del sistema. Son propiedades que no dependen ni cambian con la extensión o el tamaño del sistema. Por lo general, se obtienen dividiendo las propiedades generalizadas o las propiedades extensivas por la extensión o el tamaño de la materia para hacerlas independientes del tamaño / extensión / masa.

Los ejemplos de propiedades específicas incluyen capacidad calorífica específica, volumen específico, etc. Usualmente tienen unidades de unidades generales / unidades de masa.

Las propiedades específicas de la materia son más importantes que las generales porque

- Ayudan en las comparaciones generales de las propiedades de diferentes materiales. Se utilizan para clasificar, clasificar y comparar propiedades de diferentes materiales.

- Se utilizan en la tabla / datos de referencia para registrar fácilmente propiedades de materia fácilmente accesibles. Ayuda a registrar estándares que son generales e independientes de tamaños / extensiones / masa, manteniendo así la tabla / datos / tabla de referencia precisa y concisa.

- Nos proporcionan valores que son fáciles de memorizar y recordar, a diferencia de tratar de agrupar las diferentes propiedades de diferentes masas / tamaños de materia.

Hope this Helps!!!

¡¡¡Espero que esto ayude!!!

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Two point charges of -7uC and 4uC are a distance of 20
aivan3 [116]

Answer:

Approximately 0.979 J.

Explanation:

Assume that the two charges are in vacuum. Apply the coulomb's law to find their initial and final electrical potential energy \mathrm{EPE}.

\displaystyle \mathrm{EPE} = \frac{k \cdot q_1 \cdot q_2}{r},

where

  • The coulomb's constant k = 8.99\times 10^{9}\; \rm N\cdot m^{2} \cdot C^{-2},
  • q_1 and q_2 are the sizes of the two charges, and
  • r is the separation of (the center of) the two charges.

Note that there's no negative sign before the fraction.

Make sure that all values are in SI units:

  • q_1 = -7\rm \;\mu C = -7\times 10^{-6}\; C;
  • q_2 = 4\rm \;\mu C = 4\times 10^{-6}\; C;
  • Initial separation: \rm 20\; cm = 0.20\; cm;
  • Final separation: \rm 90\; cm = 0.90\; cm.

Apply Coulomb's law:

Initial potential energy:

\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.20}\\&= \rm -1.2586\; J\end{aligned}.

Final potential energy:

\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.90}\\&= \rm -0.279689\; J\end{aligned}.

The final potential energy is less negative than the initial one. In other words, the two particles gain energy in this process. The energy difference (final minus initial) will be equal to the work required to move them at a constant speed.

\begin{aligned}\text{Work required} &= \text{Final EPE} - \text{Initial EPE}\\&= \rm  -0.279689\; J - (-1.2586\; J)\\&\approx 0.979\; J\end{aligned}.

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I believe destructive interference
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A cheetah can run at approximately 100 km/hr and a gazelle at 80 km/hr. If both animals are running at full speed, with a gazell
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The cheetah's speed is 100x and The gazelle's speed is 80x + 70. Set the two equations equal to each other: 100x = 80x +70 (then subtract 80x from both sides). 20x = 70 (then divide by 20). X =3.5. The cheetah catches the gazelle after 3.5
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Ella makes this table to organize her notes on whether atoms gain or lose energy during the changes of state.
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The answer is D.

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The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
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