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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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Ety ratio
horrorfan [7]

3) The work done is D. zero

4) The kinetic energy is B. 180 J

5) The potential energy is A. 120 J

6) The work done depends on B. position

7) The example of non-renewable energy is C. coal

8) The power expended is 3\cdot 10^4 W

9) The efficiency is A. 100%

10) The velocity ratio is 5

Explanation:

3)

The work done by a force acting an object is given by:

W=Fd cos \theta

where :

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

When the force is applied perpendicular to the direction of motion,

\theta=90^{\circ}

Therefore, the work done is:

W=Fd(cos 90^{\circ})=0

4)

The kinetic energy of a body is given by

K=\frac{1}{2}mv^2

where

m is the mass of the body

v is its speed

For the girl in this problem, we have

m = 40 kg

v = 3 m/s

Therefore her kinetic energy is

K=\frac{1}{2}(40)(3)^2=180 J

5)

The potential energy of an object is given by

PE=mgh

where

m is the mass

g=10 m/s^2 is the acceleration of gravity

h is the heigth of the object relative to the ground

For the ball in this problem,

m = 0.4 kg

h = 30 m

So, the potential energy is

PE=(0.4)(10)(30)=120 J

6)

A conservative field is a field for which the work done by the field on an object does not depend on the path taken, but only on the initial and final position of the object.

Gravitational and electric fields are examples of conservative fields. In fact:

  • When an object is pulled down by gravity (free fall), the work done by the gravitational field only depends on the change in height \Delta h between the two points, not on the path taken during the fall
  • When an electric charge is pushed by the electric field, the work done by the field depends only on the initial and final position of the charge in the field

For any conservative field, it is possible to define a "potential" function, which represents the energy per unit mass/charge, and depends only on the position of the object.

7.

  • Non-renewable energy sources are sources of energy whose rate of consumption is faster than the rate at which they are re-created. Examples of non-renewable sources are coal, oil, natural gas. These energy sources are consumed at a fast rate, while they take million of years to regenerate, so at the current rate they will eventually run out.
  • Renewable energy sources are sources of energy that replenish at faster rate than the rate at which it is consumed. Examples of renewable sources are solar energy, wind, hydroelectric power.

Therefore, the example of non-renewable energy in this case is

C. Coal

8.

For an object pushed by a force F and moving at a constant velocity v, the power expended is given by

P=Fv

where F is the force and v is the velocity.

for the rocket in this problem, we have:

F = 10 N is the force propelling the rocket

v = 3000 m/s is its velocity

Substituting into the equation, we find the power expended:

P=(10)(3000)=30,000 W = 3\cdot 10^4 W

9.

The efficiency of a machine is given by

\eta = \frac{W_{out}}{W_{in}}

where

W_{in} is the energy in input to the machine

W_{out} is the useful work in output from the machine

For a real machine, the useful work in output is always lower than the energy input, because part of the energy is "wasted" and converted into thermal energy due to the presence of internal frictions. However, for an ideal machine, all the input energy is converted into useful work, so

W_{out}=W_{in}

And therefore the efficiency is

\eta=1

which means 100%.

10.

The velocity ratio of a block and tackle system is the ratio between the distance moved by the effort and the distance moved by the load.

VR=\frac{d_{eff}}{d_{load}}

In a block and tackle system, the velocity ratio is also equal to the number of pulleys in the system.

For the system in the problem, there are 5 pulleys: therefore, this means that when the effort moves 5 metres, the load moves 1 metres, therefore the velocity ratio is

VR=\frac{5}{1}=5

Learn more about kinetic and potential energy:

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5 0
3 years ago
How are the spiral arms of the milky way detected?
Step2247 [10]
The spiral structure emerges when galactic clusters (open), H II regions and O & B type stars (young stars) are used as tracers. We know this to be true as other pinwheel galaxies exhibit the same patterns across these tracers as in the milky way.
6 0
3 years ago
how much is need to lift a load of 100n placed at a distance of 29 cm from fulcrum if effort is applied at 60cm from the fulcrum
dangina [55]

Answer:

206.8965517 n

Explanation:

First, we need to see that 60:29 is 2.078965517:1. Then we need to multiply the energy put 29 cm from the fulcrum by 2.078965517, giving us the end result of our answer.

6 0
2 years ago
Two lasers, one red (with wavelength 633.0 nmnm) and the other green (with wavelength 532.0 nmnm), are mounted behind a 0.150-mm
Ratling [72]

(a) The screen  is 3.20m from the split.

(b) The closest minima for green, distance Δy = 0.45 cm.

When a wave hits a barrier or opening, numerous events are referred to as diffraction. It is described as the interference or bending of waves via an aperture or around the corners of an obstruction into the area that forms the geometric shadow of the obstruction or aperture.

(a)Equation of minima = sinθ  = mλ/α

Given, m = 3, λ = 6.33X10⁻⁷, α = 0.00015

Putting the values in formula to get θ.

  θ = sin⁻¹ ( \frac{3 X 6.33X10^{-7} }{0.00015} ) = 0.01266 rad

triangle need to be drawn to find relationship between θ, y$ and L

tan(θ) = y/L  where; y = 4.05 cm

L = y/tan(θ) = 3.20

Hence, the screen is 3.20m from the split.

(b) Find the closest minima for green

minima equation is sinθ  = mλ/α where, m = 4 (minima with smallest distance)

sinθ  = 4λ/α

θ = sin⁻¹ (\frac{4X6.33X10^{-7} }{0.00015}) = 0.01688 rad

Calculate L using

tanθ = y/L

  L = 4.5 cm

From equation subtract y₃ from y:

                 4.50 cm - 4.05 cm = 0.45 cm

Hence, distance Δy = 0.45 cm.

Learn more about the Diffraction with the help of the given link:

brainly.com/question/12290582

#SPJ4

I understand that the question you are looking for is "Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the ot

Question

Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the other side of the slit is a white screen. When the red laser is turned on, it creates a diffraction pattern on the screen.

a. The distance y3,red from the center of the pattern to the location of the third diffraction minimum of the red laser is 4.05 cm. How far L is the screen from the slit? Express this distance L in meters to three significant figures.

b. With both lasers turned on, the screen shows two overlapping diffraction patterns. The central maxima of the two patterns are at the same position. What is the distance Δy between the third minimum in the diffraction pattern of the red laser (from Part A) and the nearest minimum in the diffraction pattern of the green laser?

5 0
1 year ago
A circle has an initial radius of 50ft when the radius begins decreasing at a rate of 2ft/s. what is the rate of change of the a
valkas [14]
The area of the circle with radius r is
A = πr²

The rate of change of area with respect to time is
\frac{dA}{dt} = \frac{dA}{dr} . \frac{dr}{dt} =2 \pi r. \frac{dr}{dt}

The rate of change of the radius is given as
\frac{dr}{dt} =-2 \,  \frac{ft}{s}
Therefore
\frac{dA}{dt} =-4 \pi r \,  \frac{ft^{2}}{s}

When r = 10 ft, obtain
\frac{dA}{dt}|_{r=10 \, ft} = -40 \pi  \,  \frac{ft^{2}}{s}

Answer: - 40π ft²/s (or - 127.5 ft²/s)
7 0
3 years ago
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