All categories

2 years ago

Ac circuits have… A. Two current paths. B many current paths. C one current path. D two voltage paths

Physics

1 answer:

kow [346]2 years ago

3 0

I would say…

B) they have many current paths

You might be interested in

Describe the relationship between joules, meters, and newtons

Naily [24]

Newton is a unit of force. Joules is an amount of work which is equal to the force times distance, or newton meters. So the product of newtons and meters makes joules.

5 0

3 years ago

In february a little town received snow on a daily basis. The first day they had 10 inches, 3 inches the second, 7 inches the 3r

viktelen [127]

Answer:

Explanation:

Add 10+7+3+4 and then divide your answer by 4

5 0

3 years ago

Read 2 more answers

Table salt is considered a(n)<br> because it conducts electricity in water.

Lemur [1.5K]

Answer:

I'm not sure what your answer choices are but I would consider it a conductor.

Explanation: Things that conduct anything are conductors.

I hope that this will help, good luck.

7 0

3 years ago

UN ELEVADOR DE TALLER TIENE PISTONES DE ENTRADA Y DE SALIDA , CON DIAMETRO DE 15 cm Y 40 cm RESPECTIVAMENTE, SE USA EL ELEVADOR

Studentka2010 [4]

Answer:

a) La fuerza que se aplica al pistón de entrada tiene una magnitud de 1968.75 newtons.

b) La presión que se aplica al pistón de entrada es de 111408.460 pascales.

Explanation:

a) Este problema se resuelve mediante el Principio de Pascal, el cual establece que la presión dentro de un sistema hidráulico cerrado es la misma en cualquiera de sus puntos. Por tanto, podemos calcular la fuerza aplicada al pistón de entrada mediante la siguiente relación:

$\frac{F_{in}}{\frac{\pi}{4}\cdot D_{in}^{2}} = \frac{M\cdot g}{\frac{\pi}{4}\cdot D_{out}^{2} }$

$F_{in} = \left(\frac{D_{in}}{D_{out}} \right)^{2}\cdot M\cdot g$ (1)

Where:

$F_{in}$ - Fuerza aplicada al pistón de entrada, en newtons.

$D_{in}$ - Diámetro del pistón de entrada, en metros.

$D_{out}$ - Diámetro del pistón de salida, en metros.

$M\cdot g$ - Peso del carro, en newtons.

Si sabemos que $D_{in} = 0.15\,m$ , $D_{out} = 0.40\,m$ y $M\cdot g = 1.4\times 10^{4}\,N$ , entonces la fuerza aplicada al pistón de entrada es:

$F_{in} = \left(\frac{0.15\,m}{0.40\,m} \right)^{2}\cdot (1.4\times 10^{4}\,N)$

$F_{in} = 1968.75\,N$

La fuerza que se aplica al pistón de entrada tiene una magnitud de 1968.75 newtons.

b) Ahora, presión aplicada al pistón de entrada ( $p$ ), en pascales, queda descrita a través de la siguiente expresión:

$p = \frac{F_{in}}{\frac{\pi}{4}\cdot D_{in}^{2} }$ (2)

Si sabemos que $F_{in} = 1968.75\,N$ y $D_{in} = 0.15\,m$ , entonces la presión aplicada al pistón de entrada es:

$p = \frac{1968.75\,N}{\frac{\pi}{4}\cdot (0.15\,m)^{2} }$

$p = 111408.460\,Pa$

La presión que se aplica al pistón de entrada es de 111408.460 pascales.

3 0

3 years ago

Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =

Troyanec [42]

Answer:

The magnitude of the electrostatic force on $\mathrm{+32\,\mu C}$ is: $\mathbf{12\;N}$

Explanation:

Consider

$q_1=+32\;\mu\text{C}$ is at position $x_1=0\;\text{m}$
$q_2=+20\;\mu\text{C}$ is at position $x_2=40\;\text{cm}$
$q_3=-60\;\mu\text{C}$ is at position $x_3=60\;\text{cm}$

The sum of all horizontal forces on $q_1$ is given as

$\sum F_x=F_{12}+F_{13}$

where

$F_{12}$ is the force exerted by $q_2$
$F_{13}$ is the force exerted by $q_3$

The force on $q_1$ exerted by $q_2$ is repulsive, so the direction of the force $F_{12}$ is to the left (negative direction). Thus

$F_{12}=-\frac{K\,|q_1|\,|q_2|}{r_{12}^2}\\F_{12}=-\frac{K\,|q_1|\,|q_2|}{(x_2\;-\;x_1)^2}\\F_{12}=\mathrm{-\frac{(8.988\times10^9\;\frac{Nm^2}{C^2})\,|+32\;\mu C|\,|+20\;\mu C|}{(40\;cm\;-\;0\;cm)^2}}\\F_{12}=\mathrm{-\frac{(8.988\times10^9\;\frac{Nm^2}{C^2})\,(32\;\mu C)\,(20\;\mu C)}{(40\;cm)^2}}\\F_{12}=\mathrm{-\frac{(8.988\times10^9\;\frac{Nm^2}{C^2})\,(32\times10^{-6}\;C)\,(20\times10^{-6}\;C)}{(0.40\;m)^2}}\\F_{12}=\mathrm{-36\;N}$

The force on $q_1$ exerted by $q_3$ is attractive, so the direction of the force $F_{13}$ is to the right (positive direction). Thus

$F_{13}=+\frac{K\,|q_1|\,|q_3|}{r_{13}^2}\\F_{13}=+\frac{K\,|q_1|\,|q_3|}{(x_3\;-\;x_1)^2}\\F_{13}=\mathrm{+\frac{(9\times10^9\;\frac{Nm^2}{C^2})\,|+32\;\mu C|\,|-60\;\mu C|}{(60\;cm\;-\;0\;cm)^2}}\\F_{13}=\mathrm{+\frac{(9\times10^9\;\frac{Nm^2}{C^2})\,(32\;\mu C)\,(60\;\mu C)}{(60\;cm)^2}}\\F_{13}=\mathrm{+\frac{(9\times10^9\;\frac{Nm^2}{C^2})\,(32\times10^{-6}\;C)\,(60\times10^{-6}\;C)}{(0.60\;m)^2}}\\F_{13}=\mathrm{+48\;N}$

Therefore

$\sum F_x=\mathrm{-36\;N+48\;N} \;\;\;\;\;\Rightarrow\;\;\;\;\; \mathbf{\sum F_x=+12\;N}$

the electrostatic force on $q_1$ is to the right and has a magnitude of $\mathbf{12\;N}$

8 0

3 years ago