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My name is Ann [436]

2 years ago

8

Name two objects that use electric motors and are commonly found in houses.

2 answers:

yawa3891 [41]2 years ago

6 0

Answer:

There are many common household objects that feature electric motors, such as vacuum cleaners, blenders, the sink garbage disposal, and the hair dryer. There are also small electric motors in common toys, such as in the tiny cars that make them go. Some office supplies have them too: pencil sharpeners, coffee-bean grinders, ceiling fans, the printer, and even the vibrator in a cell phone!

Explanation:

<em>God Bless!</em>

DiKsa [7]2 years ago

4 0

<span>Refrigerator - Contains Induction motor in compressor.
</span><span>Toys - DC motors.</span>

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Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit.

Effectus [21]

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, $\lambda=608\ nm=608\times 10^{-9}\ m$

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

$y=\dfrac{mD\lambda}{a}$

where

a = width of the slit

$a=\dfrac{mD\lambda}{y}$

$a=\dfrac{5\times 0.885\ m\times 608\times 10^{-9}\ m}{0.0161\ m}$

a = 0.000167 m

$a=1.67\times 10^{-4}\ m$

a = 0.167 mm

So, the width of the slit is 0.167 mm. Hence, this is the required solution.

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3 years ago

A standard inverting op-amp circuit has an R1 of 10 kΩ and an Rf of 220 kΩ. If the offset current is 100 nA the output offset vo

kiruha [24]

Answer:

The value is $V_{os} = 0.001 \ V$

Explanation:

From the question we are told that

The circuit resistance is $R_1 = 10 \ k \Omega$

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Generally the offset voltage is mathematically reparented as

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=> $V_{os} = 10 *10^{3}* 100 *10^{-9}$

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Feliz [49]

The potential energy of the spring is 6.75 J

The elastic potential energy stored in the spring is given by the equation:

$E= \frac{1}{2} kx^2$

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k is the spring constant

x is the compression/stretching of the string

In this problem, we have the spring as follows:

k = 150 N/m is the spring constant

x = 0.3 m is the compression

Substituting in the equation, we get

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