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IceJOKER [234]
3 years ago
6

El tubo de entrada que suministra presión de aire para operar un gato hidráulico tiene 2 cm de diámetro. El pistón de salida es

de 32 cm de diámetro. ¿Qué presión de aire se tendrá de aire se tendrá que usar para levantar un automóvil de 17,640 N?`
Physics
1 answer:
topjm [15]3 years ago
3 0

Answer:

P₁ = 219.3 Pa

Explanation:

This fluid mechanics problem, we can use that the pressure is distributed with the same value throughout the system, which is Pascal's principle.

Let's use the subinidce1 for the small diameter and the subscript 2 for the larger diameter.

              P₁ = P₂

pressure is defined by

             P = F / A

we subtitute

             F₁ / A₁ = F₂ / A₂

             F₁ = F₂ A₁ / A₂

the area in a circle is

             A = π r² = π d² / 4

we substitute

            F₁ = F₂ (d₁ / d₂)²

we calculate

           F₁ = 17640 (2/32)²

           F₁ = 68.9 N

Having the force to be applied we can find the air pressure on the small plunger

          P₁ = F₁ / A₁

          P₁ = F₁ 4 / π d₁²

let's calculate

          P₁ = 68.9 4 / (π 0.02²)

          P₁ = 219.3 Pa

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Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee
JulsSmile [24]

Answer:

The non-relativistic kinetic energy of a proton is 6.76\times10^{-12}\ J

The relativistic kinetic energy of a proton is 7.25\times10^{-12}\ m/s

Explanation:

Given that,

Mass of proton m=1.67\times10^{-27}\ kg

Speed v= 9.00\times10^{7}\ m/s

We need to calculate the kinetic energy for non relativistic

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Put the value into the formula

K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2

K.E=6.76\times10^{-12}\ J

We need to calculate the kinetic energy for relativistic

Using formula of kinetic energy

K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)

K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)

K.E=7.25\times10^{-12}\ m/s

Hence, The non-relativistic kinetic energy of a proton is 6.76\times10^{-12}\ J

The relativistic kinetic energy of a proton is 7.25\times10^{-12}\ m/s

7 0
3 years ago
An air bubble has a volume of 2.0 cm3 when it is released by a submarine 100 m below the surface of a freshwater lake. What is t
UkoKoshka [18]

Answer:

21.35 cm^3

Explanation:

let the volume at the surface of fresh water is V.

The volume at a depth of 100 m is V' = 2 cm^3

temperature remains constant.

density of water, d = 1000 kg/m^3

Pressure at the surface of fresh water is atmospheric pressure,

P = Po = 1.013 x 10^5 N/m^2

The pressure at depth 100 m is P' = Po + hdg

P' = 1.013 \times 10^{5}+ 100 \times 1000 \times 9.8

P' = 10.813 x 10^5 N/m^2

Use the Boyle's law

P V = P' V'

1.013 \times 10^{5}\times V = 10.813 \times 10^{5}\times 2

V = 21.35 cm^3

Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.

5 0
3 years ago
a machine gun fires 10 rounds per second the speed of the bullets is 300 m/s. what is the distance in the air between the flying
vovangra [49]
<span>(300 m/s)/(10 r/s) = 30 m/round.</span>
4 0
3 years ago
Indica qué es una propiedad específica de la materia. Además explica por qué son útiles las propiedades específicas de la materi
Katyanochek1 [597]

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!!!

7 0
3 years ago
What is the magnetic flux density (B-field) at a distance of 0.36 m from a long, straight wire carrying a current of 3.8 A in ai
olga nikolaevna [1]

Answer:

The magnetic flux density is 2.11\times10^{-6}\ T

Explanation:

Given that,

Distance = 0.36 m

Current = 3.8 A

We need to calculate the magnetic flux density

Using formula of magnetic field

B =\dfrac{\mu_{0}I}{2r}

Where,

r = radius

I = current

Put the value into the formula

B =\dfrac{4\pi\times10^{-7}\times3.8}{2\times\pi\times0.36}

B=2.11\times10^{-6}\ T

Hence, The magnetic flux density is 2.11\times10^{-6}\ T

3 0
3 years ago
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