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Alenkasestr [34]
4 years ago
9

A person who chooses a chicken leg that provides 0.5 milligram of iron and 95 kcalories instead of 2 tablespoons of peanut butte

r that also provides 0.5 milligram of iron but 188 kcalories is using the principle of nutrient:a. control.b. adequacy.c. density.d. moderation.
Physics
1 answer:
Leto [7]4 years ago
5 0

Answer:

c. Density.

Explanation:

The principle of nutrient used here is Density. Here the person had two options either he could choose chicken leg or 2 tablespoon of peanut butter both provide 0.5 milligram of iron.

But since he wanted less calorie intake he chooses chicken leg as the calorie intake from peanut butter was almost twice from butter than from the chicken.  hence option is correct.

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Air flows through a nozzle at a steady rate. At the inlet the density is 2.21 kg/m3 and the velocity is 20 m/s. At the exit, the
Vitek1552 [10]

To solve the problem, it is necessary to apply the concepts related to the change of mass flow for both entry and exit.

The general formula is defined by

\dot{m}=\rho A V

Where,

\dot{m} = mass flow rate

\rho = Density

V = Velocity

Our values are divided by inlet(1) and outlet(2) by

\rho_1 = 2.21kg/m^3

V_1 = 20m/s

A_1 = 60*10^{-4}m^2

\rho_2 = 0.762kg/m^3

V_2 = 160m/s

PART A) Applying the flow equation we have to

\dot{m} = \rho_1 A_1 V_1

\dot{m} = (2.21)(60*10^{-4})(20)

\dot{m} = 0.2652kg/s

PART B) For the exit area we need to arrange the equation in function of Area, that is

A_2 = \frac{\dot{m}}{\rho_2 V_2}

A_2 = \frac{0.2652}{(0.762)(160)}

A_2 = 2.175*10^{-3}m^2

Therefore the Area at the end is 21.75cm^2

3 0
4 years ago
Which of the following is true about the mass of an object?
Butoxors [25]
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5 0
3 years ago
Is rose quartz igneous sedimentary or metamorphic
Dmitriy789 [7]
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7 0
3 years ago
A 4.00 kg block is suspended from a spring with k 500 N/m. A 50.0 g bullet is fired into the block from directly below with a sp
Yanka [14]

Answer:

a. A = 0.1656 m

b. % E = 1.219

Explanation:

Given

mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150  m/s , k = 500 N / m

a.

To find the amplitude of the resulting SHM using conserver energy

ΔKe + ΔUg + ΔUs = 0

¹/₂ * m * v²  -  ¹/₂ * k * A² = 0

A = √ mB * vₓ² / k

vₓ = mb * u₁ / mb + mB

vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518

A = √ 4.0 kg * (1.852 m/s)²   /   (500 N / m)

A = 0.1656 m

b.

The percentage of kinetic energy

%E = Es / Ek

Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5

Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N

% E = 13.72 / 1125 = 0.01219 *100

% E = 1.219

6 0
3 years ago
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