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Reika [66]
3 years ago
13

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 - L cans per min

ute. At point 2 in the pipe, the gauge pressure is 152kPa and the cross-sectional area is 8.00cm2. At point 1, 1.35m above point 2, the cross-sectional area is 2.00cm2. Part A Find the mass flow rate. M
Physics
1 answer:
Harman [31]3 years ago
4 0

Answer:

1.302\ \text{kg/s}

Explanation:

\rho = Density of water = 1\ \text{kg/L}

dV = Change in volume = 220\times 0.355\ \text{L}

dt = Time elapsed = 1 minute = 60 seconds

Mass flow rate is given by

\dot{m}=\rho\dfrac{dV}{dt}\\ =1\times \dfrac{220\times 0.355}{60}\\ =1.302\ \text{kg/s}

The mass flow rate is 1.302\ \text{kg/s}.

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Answer:

A)   F_g = 4.05 10⁻⁴⁷ N, B)   F_e = 9.2 10⁻⁸N, C)    \frac{F_e}{F_g} = 2.3 10³⁹

Explanation:

A) It is asked to find the force of attraction due to the masses of the particles

Let's use the law of universal attraction

            F = G \frac{m_1m_2}{r^2}

let's calculate

            F = 6.67 \ 10^{-11} \ \frac{9.1 \ 10^{-31} \ 1.67 \ 10 ^{-27} }{(5 \ 10^{-11})^2 }

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B) in this part it is asked to calculate the electric force

Let's use Coulomb's law

            F = k \  \frac{q_1q_2}{r^2}

let's calculate

            F = 9 \ 10^9 \  \frac{(1.6 \ 10^{-19} )^2}{(5 \ 10^{-11})^2}

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C) It is asked to find the relationship between these forces

        \frac{F_e}{F_g} = \frac{9.2 \ 10^{-8} }{4.05 \ 10^{-47} }

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therefore the electric force is much greater than the gravitational force

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M = 30.3kg<br>M = 40.17kg 9<br>R = 0.5m<br>G = 6. 67x10^11<br>F ?​
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Answer:

m¹=30.3kg

m²=40.17kg

R=0.5m

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F=Gm¹m²/R²

=160.68

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