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2 years ago

The single-pole, single-throw switch is normally wired ? between the source and the load to turn devices on and off.

Physics

1 answer:

zzz [600]2 years ago

7 0

In series.

Single-pole and single-throw switch:

A switch with only one input and one output is referred to as a Single Pole Single Throw (SPST) switch. This indicates that it has a single output terminal and a single input terminal.

A single pole, one throw switch functions as an on/off switch in circuits. The circuit is turned on when the switch is closed. The circuit is shut off when the switch is open.

Thus, SPST switches are relatively basic in design.

Circuit for a single-pole, single-throw (SPST) switch

Types:

According to the application, it can be divided into three categories, including:

Simple SPST

(ON)-OFF, Push-to-close, SPST Momentary

ON-(OFF), Push-to-Open, SPST Momentary

Inches Switch SPST

Learn more about terminal here:

brainly.com/question/14236970

#SPJ4

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he membrane that surrounds a certain type of living cell has a surface area of 6.0 x 10-9 m2 and a thickness of 1.6 x 10-8 m. As

k0ka [10]

Answer:

$1.54481175\times 10^{-12}\ C$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

A = Area = $6\times 10^{-9}\ m^2$

d = Thickness = $1.6\times 10^{-8}\ m$

k = Dielectric constant = 5.4

V = Voltage = 86.2 mV

Charge is given by

$Q=CV\\\Rightarrow Q=k\epsilon\dfrac{A}{d}V\\\Rightarrow Q=5.4\times 8.85\times 10^{-12}\times \dfrac{6\times 10^{-9}}{1.6\times 10^{-8}}\times 86.2\times 10^{-3}\\\Rightarrow Q=1.54481175\times 10^{-12}\ C$

The charge on the outer surface is $1.54481175\times 10^{-12}\ C$

4 0

3 years ago

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$

The position will be:

$x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m$

The position of the object on which its speed is one-half its maximum velocity is 0.039

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4 years ago

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3 years ago

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4 years ago

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