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3 years ago

. Using the following equation: F = ma Solve for m (1 point)

1 answer:

6 0

<span>To solve for m in the equation F = ma, you must divide both side of the equation by a. This will make the equation look like F/a = ma/a. Since m is being multiplied by a, dividing it will cancel out. Now making the final equation look like F/a=m and/or m=F/a.</span>

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Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

Temperature = $15^{o}C$ = (15 + 273) K = 288 K

Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

= $309.8 \times 10^{5} Pa$

= 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.

6 0

3 years ago

Which coefficients correctly balance the formula equation nh4no2(s)-> n2(g) + h2o(l)?

horrorfan [7]

NH4NO2 > N2 + 2 H2O

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Which of the following represents the atomic number of an element?

Anna11 [10]

Answer:

The atomic number is the number of protons in the nucleus

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jolli1 [7]

One would be phosporous whose configuration is 1s2 2s2 2p6 3s2 3p3

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nata0808 [166]

Answer:

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2 years ago