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Artyom0805 [142]
2 years ago
10

The actual mass of a sheet of copy paper is 4.5 g. What must be the actual mass of a 1 cm² plece of copy paper?

Physics
1 answer:
Allushta [10]2 years ago
3 0

The actual mass of a 1 cm² piece of copy paper is determined as 0.0074 g.

<h3>Area of the sheet of paper in square centimeters</h3>

Area = Length x Width

Area = (11 x 2.54 cm) x (8.5 x 2.54 cm)

Area = 603.22 cm²

<h3>Actual mass of the sheet of paper</h3>

Mass = area density/area

Mass = (4.5 gcm²)/(603.22 cm²)

Mass = 0.0074 g

The complete question is below:

A sheet of 8.5 inch by 11inch paper has a mass of 4.50 grams one square centimeter , 1cm2 of that sheet of paper will have a mass closets to ____grams ( (hint use 1inch = 2.54 centimeters and area = length times)

Learn more about mass here: brainly.com/question/1762479

#SPJ1

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An electromagnetic wave of wavelength 435 nm is traveling in vacuum in the —z direction. The electric field has an amplitude of
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Answer:

a) 6.9*10^14 Hz

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Explanation:

Given that

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Amplitude of the electric field, E(max) = 2.7*10^-3 V/m

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What is the specially shaped crystal that can bent light?​
suter [353]

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An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper
dimulka [17.4K]

Answer:

The power output of this engine is  P =  17.5 W

The  the maximum (Carnot) efficiency is  \eta_c  = 0.7424

The  actual efficiency of this engine is  \eta _a  = 0.46

Explanation:

From the question we are told that

    The temperature of the hot reservoir is  T_h = 1250 \ K

      The temperature of the cold reservoir  is  T_c  =  322 \ K

     The energy absorbed from the hot reservoir is E_h  = 1.37 *10^{5} \ J

       The energy exhausts into  cold reservoir is  E_c  = 7.4 *10^{4} J

The power output is mathematically represented as

      P  =  \frac{W}{t}

Where t is the time taken which we will assume to be 1 hour =  3600 s  

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The Carnot efficiency is mathematically represented as

          \eta_c  =  1 - \frac{T_c}{T_h}

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         \eta_c  = 0.7424

The actual efficiency is mathematically represented as

        \eta _a  =   \frac{W}{E_h}

substituting values

         \eta _a  =  \frac{63000}{1.37*10^{5}}

         \eta _a  = 0.46

     

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