All categories

3 years ago

A traves de una manguera de 1 in de diámetro fluye gasolina con una velocidad media de 5ft/s ¿cuál es el gasto?

Physics

1 answer:

jonny [76]3 years ago

6 0

Answer:

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

Explanation:

El gasto es el flujo volumétrico de gasolina ( $Q$ ), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:

$Q = \frac{\pi}{4}\cdot D^{2}\cdot v$ (1)

Donde:

$D$ - Diámetro de la manguera, medido en pies.

$v$ - Velocidad medida de salida, medida en pies por segundo.

Si sabemos que $D = \frac{1}{12}\,ft$ y $v = 5\,\frac{ft}{s }$ , entonces el gasto de gasolina es:

$Q = \frac{\pi}{4}\cdot \left(\frac{1}{12}\,ft \right)^{2} \cdot \left(5\,\frac{ft}{s} \right)$

$Q \approx 0.0273\,\frac{ft^{3}}{s}$

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

You might be interested in

A 300 g glass thermometer initially at 23 ◦C is put into 236 cm3 of hot water at 87 ◦C. Find the final temperature of the thermo

DIA [1.3K]

Answer:

$74^{\circ} C$

Explanation:

We are given that

Mass of glass, $m=300 g$

$T_1=23^{\circ}$

Volume, $V=236cm^3$

Mass of water= $density\times volume=1\times 236=236 g$

Density of water= $1g/cm^3$

Temperature of hot water, $T=87^{\circ}$

Specific heat of glass, $C_g=0.2cal/g^{\circ}C$

Specific heat of water, $C_w=1 cal/g^{\circ}C$

$Q_{glass}=m_gC_g(T_f-T_1)=300\times 0.2(T_f-23)$

$Q_{water}=m_wC_w(T_f-T)=236\times 1(T_f-87)$

$Q_{glass}+Q_{water}=0$

$300\times 0.2(T_f-23)+236\times 1(T_f-87)$

$60T_f-1380+236T_f-20532=0$

$296T_f=20532+1380=21912$

$T_f=\frac{21912}{296}=74^{\circ} C$

5 0

2 years ago

Match each statement with the characteristic of scientists.

Dimas [21]

Answer:

Options B, A, D, C

Explanation:

When a scientists, let's say Roberto wonders if the presence of other elements also affects the color of a flame, he can decide to prove this through a study. Therefore, in chemistry class, Roberto sees that traces of lithium makes a flame appear bright red. Subsequently, Roberto designs an experiment to test flame color in the presence of different elements and finally Roberto's friend tells him the color of a flame cannot be changed, but Roberto is still unsure.

5 0

3 years ago

Read 2 more answers

Using your Periodic Table, which element below has the smallest atomic radius? A.) Sodium, B.) Chlorine, C.) Phosphorus, D.) Iro

egoroff_w [7]

Explanation:

Chlorine is the smallest atomic radius

6 0

2 years ago

Read 2 more answers

SOMEONE HELP PLEASE ASAP I BELIEVE THE ANSWER IS C OR D

Serhud [2]

You are correct the answer would be C

hope i helps

4 0

3 years ago

An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r

bixtya [17]

1) Linear charge density of the shell: $-2.6\mu C/m$

2) x-component of the electric field at r = 8.7 cm: $1.16\cdot 10^6 N/C$ outward

3) y-component of the electric field at r =8.7 cm: 0

4) x-component of the electric field at r = 1.15 cm: $1.28\cdot 10^7 N/C$ outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

The linear charge density of the cylindrical insulating shell can be found by using

$\lambda_2 = \rho A$

where

$\rho = -567\mu C/m^3$ is charge volumetric density

A is the area of the cylindrical shell, which can be written as

$A=\pi(b^2-a^2)$

where

$b=4.7 cm=0.047 m$ is the outer radius

$a=2.7 cm=0.027 m$ is the inner radius

Therefore, we have :

$\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m$

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

$E=E_1+E_2$

where:

$E_1=\frac{\lambda_1}{2\pi r \epsilon_0}$

where

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

$E_2=\frac{\lambda_2}{2\pi r \epsilon_0}$

where

$\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m$

And this field points radially inward, because the charge is negative.

Therefore, the net field is

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C$

in the outward direction.

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

$E_y=0$

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}$

where:

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

$E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C$

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0

3 years ago