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Yakvenalex [24]
1 year ago
14

this problem occurs in petroleum firefighting when fire streams are applied to the burning liquid surface at such an angle that

hot and / or burning oil is forced over the edge of the tank. the oil then flows down the side of the tank and you are now confronted with a fire on the ground as well as the one in the tank. this best defines which problem?
Physics
1 answer:
shutvik [7]1 year ago
7 0

Petroleum has less density than water. So, when fire streams are applied to the burning liquid surface at such an angle that hot and / or burning oil is forced over the edge of the tank, the oil then flows down the side of the tank and you are now confronted with a fire on the ground as well as the one in the tank.

<h3>What are the reasons not to use fire streams during deal with  burning oil?</h3>

When you spray water on a regular fire, you're attempting to lower the flame's temperature and contain the fire.

Fuel for an oil fire is oil. Temperature and oxidizer are present in a fire (oxygen). Due to the fact that oil is lighter than water, an explosion may result if it floats above water while water sinks beneath oil.

Consequently, water shouldn't be used to put out oil fires.

Learn more about density here:

brainly.com/question/15164682

#SPJ1

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An air compressor compresses 6 L of air at 120 kPa and 22°C to 1000 kPa and 400°C. Determine the flow work, in kJ/kg, required b
Mariana [72]

Answer:

The work flow required by the compressor = 100.67Kj/kg

Explanation:

The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .

The work flow can be determined using the equation:

M1h1 + W = Mh2

U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2

Workflow = P2alpha2 - P1alpha1

Workflow = (h2 -U2) - (h1 - U1)

Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)

Workflow = ( 193.191 - 92.519)Kj/kg

Workflow = 100.672Kj/kg

6 0
4 years ago
An open 1-m-diameter tank contains water at a depth of 0.5 m when at rest. As the tank is rotated about its vertical axis the ce
Morgarella [4.7K]

Answer:

Angular velocity (w) = 8.86 rad/s

Explanation:

Angular velocity (w) = \sqrt{} 4ghi/R^{2}

g= 9.81 m/s

R= 0.5

hi (initial depth) = 0.5m

Hence= \sqrt4* 9.81* 0.5/0.5^{2}  = 8.86 rad/s

3 0
3 years ago
Which formulas have been correctly rearranged to solve for radius? Check all that apply. r = GM central/v^2 r =fcm/v^2 r =ac/v^2
jek_recluse [69]

The orbital radius is: r=\frac{GM}{v^2}

Explanation:

The problem is asking to find the radius of the orbit of a satellite around a planet, given the orbital speed of the satellite.

For a satellite in orbit around a planet, the gravitational force provides the required centripetal force to keep it in circular motion, therefore we can write:

\frac{GMm}{r^2}=m\frac{v^2}{r}

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the satellite

r is the radius of the orbit

v is the speed of the satellite

Re-arranging the equation, we find:

\frac{GM}{r}=v^2\\r=\frac{GM}{v^2}

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

7 0
3 years ago
Read 2 more answers
a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)

We have

\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}

\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}

Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

5 0
2 years ago
3. In which activity is no work done?
Firlakuza [10]

Answer:

Sleeping, since your brain is working, but your physical body is not moving at all or moving rarely, therefore there is no work done in the activity of sleeping.

Explanation:

6 0
4 years ago
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