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Kryger [21]
3 years ago
10

To reduce the global emission of greenhouse gases, which of the following would be theMOST effective and practical lifestyle cha

nge for most Americans?A. Opt for polystyrene (Styrofoam) food containers instead of clear plastic.B. Drive an electric vehicle.C. Decrease your consumption of beef.D. Carry your groceries in a reusable canvas bag. - C. Decrease your consumption of beef
Engineering
1 answer:
Anestetic [448]3 years ago
4 0

Answer:

C. Decrease your consumption of beef.

Explanation:

The production of beef needs lots of land, which leads to trees being cut down, releasing carbon dioxide. Also, the ruminant animals used to produce beef end up farting a lot from their diets, and these farts are of high metane, which leads to greenhouse gases being emitted in the athmosphere.

Common plant proteins, otherwise, produce way less of these gases.

So the correct answer is:

C. Decrease your consumption of beef.

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A power washer is being used to clean the siding of a house. Water enters at 20 C, 1 atm, with a velocity of 0.2 m/s .A jet of w
BabaBlast [244]

A power washer is being used to clean the siding of a house. Water at the rate of 0.1 kg/s enters at 20°c and 1 atm, with the velocity 0.2m/s. The jet of water exits at 23°c, 1 atm with a velocity 20m/s at an elevation of 5m. At steady state, the magnitude of the heat transfer rate from power unit to the surroundings is 10% of the power input. Determine the power input to the motor in kW.

Answer:

Net power of 1.2 KW is being extracted

Explanation:

We are given;

Mass flow rate; m' = 0.1kg/s

Inlet temperature; T1 = 20°C = 293K

Inlet pressure; P1 = 1 atm = 10^(5) pa

Inlet velocity; v1 = 0.2 m/s

Exit Pressure; P2 = 1 atm = 10^(5) pa

Exit Temperature; T2 = 1 atm = 296K

Exit velocity; V2 = 20m/s

Change in elevation; h = Z2 - Z1 = 5m

We are told that the magnitude of the heat transfer rate from the power unit to the surroundings is 10% of the power input.

Thus;

Q = -0.1W

From Bernoulli equation;

Q - W = ∆Potential energy + ∆Kinetic energy + ∆Pressure energy

Where;

∆Potential energy = mg(z2 - z1)

∆Kinetic energy = ½m(v2² - v1²)

∆Pressure energy = mc_p(T2 - T1)

Thus;

-0.1W - W = [m'g(z2 - z1)] + [½m'(v2² - v1²)] + [m'c_p(T2 - T1)]

Where C_p is specific heat capacity of water = 4200 J/Kg.k

Plugging in the relevant values, we have;

-1.1W = (0.1 × 9.81 × 5) + (½ × 0.1(20² - 0.2²)) + (0.1 × 4200 × (296 - 293))

-1.1W = 4.905 + 19.998 + 1260

-1.1W = 1284.903

W = -1284.903/1.1

W ≈ -1168 J/s ≈ -1.2 KW

The negative sign means that work is extracted from the system.

7 0
3 years ago
A boiler is used to heat steam at a brewery to be used in various applications such as heating water to brew the beer and saniti
Natalija [7]

Answer:

net boiler heat = 301.94 kW

Explanation:

given data

saturated steam = 6.0 bars

temperature = 18°C

flow rate = 115 m³/h = 0.03194 m³/s

heat use by boiler = 90 %

to find out

rate of heat does the boiler output

solution

we can say saturated steam is produce at 6 bar from liquid water 18°C

we know at 6 bar from steam table

hg = 2756 kJ/kg

and

enthalpy of water at 18°C

hf = 75.64 kJ/kg

so heat required for 1 kg is

=hg - hf

= 2680.36 kJ/kg

and

from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg

so here mass flow rate is

mass flow rate = \frac{0.03194}{0.315}

mass flow rate m = 0.10139 kg/s

so heat required is

H = h × m  

here h is heat required and m is mass flow rate

H = 2680.36  × 0.10139

H =  271.75 kJ/s = 271.75 kW

now 90 % of boiler heat is used for generate saturated stream

so net boiler heat = \frac{H}{0.90}

net boiler heat = \frac{271.75}{0.90}

net boiler heat = 301.94 kW

5 0
3 years ago
What do auto mechanics do when they take out the spark plugs
ivanzaharov [21]
Check them to see if the cylinder is running rich or lean. This can be determined by looking at the electrodes on the plugs
Hope this helps.
7 0
3 years ago
A dipstick is (a direct,an indirect) measurement device
icang [17]
Direct because it’s going right in the spot it needs to be in
3 0
4 years ago
E total kinetic energy of a 2500 lbm car when it is moving at 80 mph (in BTU)?
Galina-37 [17]

Answer:

KE= 687.21 BTU

Explanation:

Given that

Mass of car = 2500 lbm

We know that 1 lb=0.45 kg

So the mass of car  m =1133.98 kg

Velocity of car= 80 mph

We know that 1 mph =0.44 m/s

So velocity of car = 35.76 m/s

As we know that kinetic energy (KE) is given as follows

KE=\dfrac{1}{2}mv^2

Now by putting the values

KE=\dfrac{1}{2}\times 1133.98\times 35.76^2

KE=725.05 KJ

We know that   1 KJ = 0.94 BTU

So  KE= 687.21 BTU

3 0
4 years ago
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