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trasher [3.6K]
2 years ago
8

A certain liquid has a density of 2.67 g/cm3. 1340 g of this liquid would occupy a volume of ________ l.

Physics
2 answers:
qaws [65]2 years ago
8 0
Volume = mass/density = 1340/2.67 = 501.87 cm^3 = 0.50187L
ch4aika [34]2 years ago
6 0

Answer: 0.5018L

Explanation: The relationship between mass and density is given by the following equation:

Volume = Mass/Density

The density given shows that 2.67g of the liquid would occupy a cm³ volume at every point in time.

Given this, total volume occupied would be

Mass = 1340g

Density = 2.67g/cm³

V =?

V = 1340/2.67

V = 501.87cm³ = 501.87/1000 = 0.5018L

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The δe of a system that releases 12.4 j of heat and does 4.2 j of work on the surroundings is __________ j.
Vedmedyk [2.9K]

The answer for this problem is clarified through this, the system is absorbing (+). And now see that it uses that the SURROUNDINGS are doing 84 KJ of work. Any time a system is overshadowing work done on it by the surroundings the sign will be +. So it's just 12.4 KJ + 4.2 = 16.6 KJ.

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2 years ago
Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is
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Answer:

T_{out}=27.0000077 ºC

Explanation:

First, let's write the energy balance over the duct:

H_{out}=H_{in}+Q

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

m*Cp(T_{out}-T_{in})=Q

So, let's isolate T_{out}:

T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}

The Cp of the air at 27ºC is 1007\frac{J}{kgK} (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are T_{out} and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

P=2*H+2*A=2*0.004m+2*0.016m=0.04m

Surface area:

A=P*L=0.04m*1m=0.04m^2

Then, the heat Q is:

600\frac{W}{m^2} *0.04m^2=24W

Finally, find the exit temperature:

T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077

T_{out}=27.0000077 ºC

The temperature change so little because:

  • The mass flow is so big compared to the heat flux.
  • The transfer area is so little, a bigger length would be required.
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An airliner of mass 1.70×105kg1.70×105kg lands at a speed of 75.0 m/sm/s. As it travels along the runway, the combined effects o
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Answer:

The airliner travels 1.65 km along the runway before coming to a halt.

Explanation:

Given

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Mass of the airliner = (1.70 × 10⁵) kg = 170000 kg

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According to the work-energy theorem, the work done by the resistive forces in stopping the airliner is equal to the travelling kinetic energy of the airliner.

Work done by the resistive forces = (290000) × d = (290,000d) J

Kinetic energy of the airliner = (1/2)(170000)(75²) = 478,125,000 J

290000d = 478,125,000

d = (478,125,000/290,000)

d = 1648.7 m = 1.65 km

Hope this helps!!!

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Answer:

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