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Vitek1552 [10]
3 years ago
11

The article provides information by using a list. What does it list? A. Thanksgiving food B. places where clams can be found C.

birds served at the first Thanksgiving D. four men who went on a "fowling" mission
Engineering
1 answer:
Gelneren [198K]3 years ago
7 0

Answer:

C

Explanation:

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The demand schedules for Jones, Smith, and other buyers are shown in the table below:
Solnce55 [7]

The demand curve is the graphical representation of the relationship between the price of a good and the quantity demanded for a given period of time.

<h3>What is a demand schedule?</h3>

A demand schedule is a table which shows the quantity demanded of a good or service at different price levels.

A demand schedule can be graphed as a continuous demand curve on a chart where the Y-axis represents the price and the X-axis represents quantity.

Here, a typical representation, the price will appear on the left vertical axis, the quantity demanded on the horizontal axis.

Note that the complete information wasn't found and an overview was given.

Learn more about demand on:

brainly.com/question/1245771

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4 0
2 years ago
A small vehicle is powered by a pulsejet. The available net thrust is 6000 N and the traveling speed is 200 km/hr. The gases lea
kumpel [21]

Answer:

a) The mass flow rate is 19.71 kg/s

b) The inlet area is 0.41 m²

c) The thrust power is 333.31 kW

d) The propulsive efficiency is 26.7%

Explanation:

Please look at the solution in the attached Word file.

Download docx
4 0
3 years ago
The conditions at the beginning of compression in an Otto engine operating on hot-air standard with k=1.35 and 101.325 kPa, 0.05
Katyanochek1 [597]

Answer:

P_m=181.42 KPa

Explanation:

P_1=101.325 KPa,V_1=0.05m^3,T_1=32C,K=1.35

Clearance is 8%.

Heat added=15 KJ

We know that compression ratio r=1+\dfrac{1}{C}

r=1+\dfrac{1}{0.08}  

r=13.5

r=\dfrac{V_1}{V_2}

13.5=\dfrac{0.05}{V_2}

V_2=3.7\times 10^{-3}

We know that efficiency of otto cycle

\eta =1-\dfrac{1}{r^{k-1}}

\eta =1-\dfrac{1}{13.5^{1.35-1}}

\eta =0.56

\eta =\dfrac{W}{Q}

W is the work out put and Q is the heat addition.

0.56 =\dfrac{W}{15}

W=8.4 KJ

We know that Work =Mean effective pressure x swept volume.

Here swept volume V_s=V_1-V_2

V_s=0.05-3.7\times 10^{-3}

V_s=0.0463 m^3

Noe by putting the values

Work =Mean effective pressure x swept volume.

8.4=P_m\times 0.0463

P_m=181.42 KPa

6 0
3 years ago
Stainless steel ball bearings (rho = 8085 kg/m3 and cp = 0.480 kJ/kg·°C) having a diameter of 1.2 cm are to be quenched in water
sveticcg [70]

Answer:

\dot Q = -2.341\,kJ

Explanation:

The rate of heat transfer from the balls to the air is:

\dot Q = \left[(800\,\frac{1}{min} )\cdot (\frac{1\,min}{60\,s} )\cdot(8085\,\frac{kg}{m^{3}})\cdot (\frac{4}{3}\pi )\cdot (6\times 10^{-3}\,m)^{3}\right]\cdot (0.480\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (850\,^{\textdegree}C-900\,^{\textdegree}C)\dot Q = -2.341\,kJ

3 0
4 years ago
Read 2 more answers
Example 1: A six-lane urban freeway (three lanes in each direction) is on level terrain with 11-ft lanes, obstructions 2 ft from
MrMuchimi

Answer:

Please find attached solution

Explanation:

5 0
3 years ago
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