Omplete the following program: [0.5 X 4 = 2]

Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of poun

RoseWind [281]

Answer:

Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of pounds), and let y be the miles per gallon (mpg)

Explanation:

6 0

3 years ago

Air is contained in a vertical piston–cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a p

oee [108]

Answer:

a) 24 kg

b) 32 kg

Explanation:

The gauge pressure is of the gas is equal to the weight of the piston divided by its area:

p = P / A

p = m * g / (π/4 * d^2)

Rearranging

p * (π/4 * d^2) = m * g

m = p * (π/4 * d^2) / g

m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg

After the weight is added the gauge pressure is 2.8kPa

The mass of piston plus addded weight is

m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg

56 - 24 = 32 kg

The mass of the added weight is 32 kg.

5 0

3 years ago

What does this work for

Anastaziya [24]

Answer:

it allows your dash board to light up you MPH RPM and all the other numbers on the spadomter

Explanat

8 0

3 years ago

Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value

masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

$q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} } )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}$

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

$q_{n} =uK(\frac{g(\rho_{L}-\rho _{v}) }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4$

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0

3 years ago

Assume that the water temperature is 10°C and the depth of the settling tank is 3.0 m (9.80 ft). Calculate the theoretical settl

Lelu [443]

Explanation:

See attached file

6 0

3 years ago