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3 years ago

Which of the following describe an electrical motor? Check all that apply

Physics

1 answer:

weeeeeb [17]3 years ago

5 0

Answer:

changes electrical energy into mechanical energy

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An objects mechanical energy never changes it simply changes between

tatiyna

Actually moving and not. It is the sum of potential and kinetic energy.

4 0

3 years ago

Can somebody help me with this

Ivanshal [37]

this is simple the word your looking for is 8 letters long and the definition is key to the answer

6 0

4 years ago

A biconvex lens is formed by using a piece of plastic(n=1.70).

creativ13 [48]

Answer:

$f =17.15\ cm$

Explanation:

given,

refractive index of lens, n = 1.70

Radius of curvature of front surface. R₁ = 20 cm

Radius of curvature of the back surface, R₂ = 30 cm

focal length= ?

$\dfrac{1}{f}=(n-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})$

R₁ = +20 cm

R₂ = -30 cm

n = 1.70

$\dfrac{1}{f}=(1.70-1)(\dfrac{1}{20}-\dfrac{1}{-30})$

$\dfrac{1}{f}=0.70 \times 0.0833$

$f = \dfrac{1}{0.7 \times 0.0833}$

$f =17.15\ cm$

the focal length of the lens is equal to 17.15 cm

3 0

3 years ago

What temperature will 1L of H20 at 200°F become when a piece of copper,0.25kg at 260.928K, comes into contact with water?

Lady_Fox [76]

Answer:

Explanation:

mass of 1 L water = 1 kg .

200⁰F = (200 - 32) x 5 / 9 = 93.33⁰C .

260.928 K = 260.928 - 273 = - 12.072⁰C .

water is at higher temperature .

Let the equilibrium temperature be t .

Heat lost by water = mass x specific heat x fall of temperature

= 1 x 4.2 x 10³ x ( 93.33 - t )

Heat gained by copper

= .25 x .385 x 10³ x ( t + 12.072 )

Heat lost = heat gained

1 x 4.2 x 10³ x ( 93.33 - t ) = .25 x .385 x 10³ x ( t + 12.072 )

93.33 - t = .0229 ( t + 12.072)

93.33 - t = .0229 t + .276

93.054 = 1.0229 t

t = 90.97⁰C .

7 0

3 years ago

Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.

olga2289 [7]

Answer:

$NO_{1.499}$

Explanation:

Let assume that 100 kg of the compound is tested. The quantity of kilomoles for each element are, respectively:

$n_{N} = \frac{36.86\,kg}{14.006\,\frac{kg}{kmol} }$

$n_{N} = 2.632\,kmol$

$n_{O} = \frac{63.14\,kg}{15.999\,\frac{kg}{kmol} }$

$n_{O} = 3.946\,kmol$

Ratio of kilomoles oxygen to kilomole nitrogen is:

$n^{*} = \frac{3.946\,kmol}{2.632\,kmol}$

$n^{*}= 1.499$

It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.

The empirical formula for the compound is:

$NO_{1.499}$

8 0

4 years ago

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