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Gnoma [55]
3 years ago
12

Buttery popcorn contained in a large 1__bowl has a mass of about 50 __and about 650 calories.​

Physics
2 answers:
Fudgin [204]3 years ago
6 0

Answer:

Buttery popcorn contained in a large 1 <u>liter</u> bowl has a mass of about 50 <u>grams</u> of fat and about 650 calories

Explanation:

The question is about using correct units. The capacity of a bowl must be measured in a volume unit; the options were milliliter and liter. 1 milliliter is a very small amount (about 0.002 pint), so 1 liter is the correct option.

For the second blank, the options (mass units) are: microgram, milligram, gram, kilogram. Milligram is used to measure the mass of minerals, microgram is even smaller than milligram, the buttered popcorn weights much less than 50 kilograms. So, the correct option is 50 grams.

Rasek [7]3 years ago
4 0

Answer:

Buttery popcorn contained in a large 1 liter bowl has a mass of about 50 mg and about 650 calories.

Explanation:

Liter is the most appropriate unit to measure a bowl. Usually 1 liter of liquid has a mass of 1000 gram. Since popcorn is something lightweight and only a few can fill the bowl quickly so 50 mg makes perfect sense with 1 liter of bowl and 650 calories in buttery popcorn.

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At the melting point
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Find the electron and hole concentrations and fermi level in silicon at 300 k for boron
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To find the electron and hole concentrations and fermi levels in silicon at 300 k for boron, fermi level is 9.3 x 10^4 cm^-3.

<h3>Calculation and Explanation</h3>

Ionization energy for boron in Si, 0.045 eV

Fermi level in silicon,

300 K at 10^{15} cm^{-3} (Boron atoms)

All boron impurities are ionized, 300K

Electron, hole concentrations and Fermi levels is found out,

The value of Na = 10^{15} cm^{-3}

np = \frac{ni^2}{nA}

np = \frac{(9.65 x 10^9)^2}{10^15}

np= 9.3 x 10^4 cm^-3

Fermi level is calculated,

Ef - Ev = kT ln(NV/ND)

value of kT = 0.0259 eV (300° K)

So, substitution the values

Ef - Ev = kT ln(NV/ND)

Ef - Ev = 0.0259ln (2.66 x 10^{19} / 10^{15} )

Ef - Ev = 0.0259ln (26600)

Ef - Ev = 0.0259 x 10.18

Ef - Ev = 0.263 eV

What is ionization energy?

Ionization energy, also known as ionization energy (IE) or ionization energy (British English spelling), is the minimal amount of energy needed to free the most loosely bonded electron from an isolated gaseous atom, positive ion, or molecule in physics and chemistry.

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brainly.com/question/16243729

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4 0
1 year ago
What is the distance in m between lines on a diffraction grating that produces a second-order maximum for 775-nm red light at an
8090 [49]

Answer:

The distance is d =  1.747 *10^{-6} \ m  

Explanation:

From the question we are told that  

       The order of maximum diffraction is  m =  2

         The wavelength is   \lambda  =  775 nm  = 775 * 10^{-9}  \ m

         The angle is  \theta =  62.5^o

Generally the   condition for  constructive  interference for diffraction grating  is mathematically represented as

          dsin \theta  =  m *  \lambda

where  d is  the distance between the lines on a  diffraction grating

     So  

            d =  \frac{m *  \lambda  }{sin (\theta  )}

substituting values  

           d =  \frac{2  *  775 *1^{-9} }{sin ( 62.5  )}

          d =  1.747 *10^{-6} \ m

   

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3 years ago
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Answer:

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A 0.70-kg basketball dropped on a hardwood floor rises back up to 65% of its original height. (a) If the basketball is dropped f
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Answer:

Part a)

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Part b)

Loss = 0.99 J

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This is loss in terms of thermal energy due to collision with the floor

Explanation:

Part a)

Since we know that the ball rises up by 65% of initial height

so after first bounce it will lose 35% of its initial energy

so we will have

U = mgH

Energy Loss = 0.35 mgH[/tex]

Loss = 0.35(0.70)(9.81)(1.5)

Loss = 3.6 J

Part b)

Energy of the ball after first bounce

U_1 = 0.65 mgH

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U_2 = 0.65(0.65 mgH)

energy of the ball after 3rd bounce

U_3 = (0.65)(0.65^2)mgH

U_3 = 0.65^3(0.70)(9.81)(1.5)

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Loss = 0.35 U_3

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Loss = 0.99 J

Part C)

This is loss in terms of thermal energy due to collision with the floor

7 0
3 years ago
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