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kenny6666 [7]
3 years ago
13

Vince is trying to figure out the volume of two mystery matters. The volume of one of the substances needs to be measured by sub

merging it in water and the other needs to be measured using a graduated cylinder. Based on the properties of two mystery matters, what are they? (2 points)
Group of answer choices

A rock and orange juice

Helium and a golf ball

Lemonade and milk

Orange juice and helium
Engineering
1 answer:
pshichka [43]3 years ago
8 0

Answer:

Rock and orange juice

Explanation:

The mystery matter to be submerged in water must be a solid, therefore we can eliminate the Lemonade and Milk, and Orange juice and Helium, as these pairs do not contain solids. The graduated cylinder is used to measure the volume of a liquid, therefore the only remaining option is Rock and Orange Juice.

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laila [671]

Answer:

When the expenditure increased, then the consumer's expenditure is increased to Rs.150 and when the price falls of the good it becomes Rs.5. Then, Good X will be Rs.10.

last question:If the Good X falls by 20%, then, it will be Rs.2, and according to his demand 100 units will be equal to Rs.200.

cause if one unit=rs.2, then 100units=100×2=200.

8 0
3 years ago
The welding method that requires the operator to observe and only make corrections is
SVEN [57.7K]
Automatic manual semiautomatic
3 0
3 years ago
Read 2 more answers
11–17 A long, thin-walled double-pipe heat exchanger with tube and shell diameters of 1.0 cm and 2.5 cm, respectively, is used t
lana [24]

Answer:

the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C

Explanation:

Given:

d₁ = diameter of the tube = 1 cm = 0.01 m

d₂ = diameter of the shell = 2.5 cm = 0.025 m

Refrigerant-134a

20°C is the temperature of water

h₁ = convection heat transfer coefficient = 4100 W/m² K

Water flows at a rate of 0.3 kg/s

Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?

First at all, you need to get the properties of water at 20°C in tables:

k = 0.598 W/m°C

v = 1.004x10⁻⁶m²/s

Pr = 7.01

ρ = 998 kg/m³

Now, you need to calculate the velocity of the water that flows through the shell:

v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2}  }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2}  }{4}) } =0.729m/s

It is necessary to get the Reynold's number:

Re=\frac{v_{w}(d_{2}-d_{1}) }{v} =\frac{0.729*(0.025-0.01)}{1.004x10^{-6} } =10891.4343

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

Nu=0.023Re^{0.8} Pr^{0.4} =0.023*(10891.4343)^{0.8} *(7.01)^{0.4} =85.0517

The overall heat transfer coefficient:

Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} }  }

Here

h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C

Substituting values:

Q=\frac{1}{\frac{1}{4100}+\frac{1}{3390.7278}  } =1855.8923W/m^{2} C

5 0
4 years ago
A cylindrical tank is required to contain a gage pressure 560 kPa . The tank is to be made of A516 grade 60 steel with a maximum
adoni [48]

Answer:

5.6 mm

Explanation:

Given that:

A cylindrical tank is required to contain a:

Gage Pressure P = 560 kPa

Allowable normal stress \sigma = 150 MPa = 150000 Kpa.

The inner diameter of the tank = 3 m

In a closed cylinder  there exist both the circumferential stress and the longitudinal stress.

Circumferential stress \sigma = \dfrac{pd}{2t}

Making thickness t the subject; we have

t = \dfrac{pd}{2* \sigma}

t = \dfrac{560000*3}{2*150000000}

t = 0.0056 m

t = 5.6 mm

For longitudinal stress.

\sigma = \dfrac{pd}{4t}

t= \dfrac{pd}{4*\sigma }

t = \dfrac{560000*3}{4*150000000}

t = 0.0028  mm

t = 2.8 mm

From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value  with the maximum thickness = 5.6 mm

8 0
3 years ago
A 4-stroke Diesel engine with a displacement of Vd = 2.5x10^-3m^3 produces a mean effective pressure of 6.4 bar at the speed of
yKpoI14uk [10]

Answer:

The power developed by engine is 167.55 KW

Explanation:

Given that

V_d=2.5\times 10^{-3} m^3

Mean effective pressure = 6.4 bar

Speed = 2000 rpm

We know that power is the work done per second.

So

P=6.4\times 100\times 2.5\times 10^{-3}\times \dfrac{2\pi \times2000}{120}

We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.

P=167.55 KW

So the power developed by engine is 167.55 KW

4 0
3 years ago
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