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Irina18 [472]
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
Physics
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 41.0 g marble moving at 2.30 m/s strikes a 25.0 g marble at rest. What is the speed of each marble immediately after the colli
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Answer:

speed of each marble after collision will be 1.728 m/sec

Explanation:

We have given mass of the marble m_1=41gram=0.041kg

Velocity of marble v_1=2.30m/sec

Its collides with other marble of mass 25 gram

So mass of other marble m_2=25gram=0.025kg

Second marble is at so v_2=0m/sec

We have to find the velocity of second marble

From momentum conservation we know that

m_1v_1+m_2v_2=(m_!+m_2)v, here v is common velocity of both marble after collision

So 0.041\times 2.30+0.025\times 0=(0.041+0.025)v

v = 1.428 m /sec

So speed of each marble after collision will be 1.728 m/sec

6 0
3 years ago
Who was the first to hypothesize that electron orbit a positively charged nucleus
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5 0
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Dante is leading a parade across the main street in front of city hall. Starting at city hall, he marches the parade 4 blocks ea
Ipatiy [6.2K]

Answer:

The correct option is A)

Displacement: 6.71 m, Direction: 63.4 degrees north of east

Explanation:

Given that Dante is leading a parade across the main street in front of city hall.

Let, Initial location of parade is 0i+0j

One block of city is one units on the XY- graph

Statement 1: Parade marches the parade 4 blocks east, then 3 blocks south

New location of parade is 4i-3j

Statement 2: The parade marches 1 block west and 9 blocks north and finally stops.

Final location of parade is (4i-3j)+(-1i+9j)=3i+6j

Displacement is given by

Displacement = (Final destination)-(Initial destination)

Displacement = (3i+6j)-(0i+0j)=3i+6j

Thus,

Magnitude of displacement = \sqrt{3^{2}+6^{2}}

                                              = 6.71 m

Direction of displacement =  tan^{-1}(\frac{Y}{X} )

                                           =  tan^{-1}(\frac{6}{3} )

                                           = 63.43 NE

Therefore, the correct option is A) Displacement: 6.71 m, Direction: 63.4 degrees north of east

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