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3 years ago

5

An on-farm diesel fuel tank is on a stand 5 feet above the ground. The tank is vented to the atmosphere (the air in the tank is

at atmospheric pressure). If the diesel level in the tank is 2.1 feet above the release point. If the density of diesel is 6.98 pounds/gallon, what is the gauge pressure of the fuel at the nozzle? What is the absolute pressure?
Gauge Pressure = XXXX psi

Absolute Pressure = XXXX psi

1 answer:

Maru [420]3 years ago

7 0

Answer:

Pg= 0.766 psi

Pabs= 15.27 psi

Explanation:

Given

h= 2.1 ft

1 ft = 0.3048 m

2.1 ft = 0.64 m

h= 0.64 m

ρ=6.98 pounds/gallon

1 pounds/gallon = 119.82 kg/m³

ρ = 825.6 kg/m³

The gauge pressure Pg

Pg= ρ g h

Pg= 825.6 x 10 x 0.64 ( take g =10 m/s²)

Pg= 5283.84 Pa

We know that

1 Pa= 0.000145 psi

Pg= 0.766 psi

Lets take atmospheric pressure Patm

Patm= 100 KPa

Pg= 5.283 KPa

Absolute pressure = Atmospheric pressure + Gauge pressure

Pabs= 100 + 5.283

Pabs= 105.283 KPa

We know that

1 KPa= 0.145 Psi

Pabs= 15.27 psi

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A cubical box measuring 1.29 m on each side contains a monatomic ideal gas at a pressure of 2.0 atm How much thermal energy do t

Marrrta [24]

Answer:

a) $U = 652.545\,kJ$ , b) $v \approx 659.568\,\frac{m}{s}$

Explanation:

a) According to the First Law of Thermodinamics, the system is not reporting any work, mass or heat interactions. Besides, let consider that such box is rigid and, therefore, heat contained inside is the consequence of internal energy.

$Q = U$

The internal energy for a monoatomic ideal gas is:

$U = \frac{3}{2} \cdot n \cdot R_{u} \cdot T$

Let assume that cubical box contains just one kilomole of monoatomic gas. Then, the temperature is determined from the Equation of State for Ideal Gases:

$T = \frac{P\cdot V}{n\cdot R_{u}}$

$T = \frac{(202.65\,kPa)\cdot(1.29\,m)^{3}}{(1\,kmole)\cdot(8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K} )}$

$T = 52.325\,K$

The thermal energy contained by the gas is:

$U = \frac{3}{2}\cdot (1\,kmole)\cdot (8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K})\cdot (52.325\,K)$

$U = 652.545\,kJ$

b) The physical model for the cat is constructed from Work-Energy Theorem:

$U = \frac{1}{2}\cdot m_{cat} \cdot v^{2}$

The speed of the cat is obtained by isolating the respective variable and the replacement of every known variable by numerical values:

$v = \sqrt{\frac{2 \cdot U}{m_{cat}}}$

$v = \sqrt{\frac{2\cdot (652.545 \times 10^{3}\,J)}{3\,kg} }$

$v \approx 659.568\,\frac{m}{s}$

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3 years ago

Which is a characteristic of nuclear funsion?

disa [49]

Nuclear fusion is where two Nucluir collide and form a new atom and release a tremendous amount of energy it requires a high activation energy and we cant use it for an energy source as its to dangerous

hope that helps

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4 years ago

A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the sp

Anna71 [15]

$\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}$

Explanation:

Natural length of a spring is $0.5\text{ }m$ . The spring is streched by $0.02\text{ }m$ . The resultant energy of the spring is $0.5\text{ }J$ .

The potential energy of an ideal spring with spring constant $k$ and elongation $x$ is given by $\dfrac{1}{2}kx^{2}$ .

So, in the current problem, the natural length of the spring is not required to find the spring constant $k$ .

$\text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}$

∴ The spring constant of the spring = $2500\text{ }\frac{kg}{s^{2}}$

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4 years ago

When energy changes form, energy is...<br><br>A) lost.<br>B) created.<br>C) not lost or created.

devlian [24]

Answer:

C

Explanation:

Energy is neither created or destroyed, but it can change forms...

6 0

3 years ago

Read 2 more answers

A mass of 150 g stretches a spring 1.568 cm. If the mass is set in motion from its equilibrium position with a downward velocity

nadezda [96]

Answer:

$u(t)=\frac{1}{5} sin\ (25t)$

Explanation:

Given:

mass of the body stretching the spring, $m=150\ g$

extension in spring, $\Delta x=1.568\ cm$

velocity of oscillation, $u'(0)=20\ cm.s^{-1}$

initial displacement position of equilibrium, $u(0)=0$

<u>According to given:</u>

$m.g=k.\Delta x$

$150\times 980=k\times 1.568$

$k=93750\ dyne.cm^{-1}$

<u>we know frequency:</u>

$\omega=\sqrt{\frac{k}{m} }$

$\omega=\sqrt{\frac{93750}{150} }$

$\omega=25$

Now, for position of mass in oscillation:

$u= A.sin\ (\omega.t)+B.cos\ (\omega.t)$

$u= A.sin\ (25.t)+B.cos\ (25.t)$

at $t=0;\ u(0)=0\ \Rightarrow A=0$

∴ $u(t)=B.sin\ (25.t)$

∵ at $t=0;\ u'(0)=20\ cm.s^{-1}\ \Rightarrow B=\frac{1}{5}$

$u(t)=\frac{1}{5} sin\ (25t)$

7 0

3 years ago