Answer:
16.2 cents
Explanation:
Given that a homeowner consumes 260 kWh of energy in July when the family is on vacation most of the time.
Where Base monthly charge of $10.00. First 100 kWh per month at 16 cents/kWh. Next 200 kWh per month at 10 cents/kWh. Over 300 kWh per month at 6 cents/kWh.
For the first 100 kWh:
16 cent × 100 = 1600 cents = 16 dollars
Since 1 dollar = 100 cents
For the remaining energy:
260 - 100 = 160 kwh
10 cents × 160 = 1600 cents = 16 dollars
The total cost = 10 + 16 + 16 = 42 dollars
Note that the base monthly of 10 dollars is added.
The cost of 260 kWh of energy consumption in July is 42 dollars
To determine the average cost per kWh for the month of July, divide the total cost by the total energy consumed.
That is, 42 / 260 = 0.1615 dollars
Convert it to cents by multiplying the result by 100.
0.1615 × 100 = 16.15 cents
Approximately 16.2 cents
The number of hectares of each crop he should plant are; 250 hectares of Corn, 500 hectares of Wheat and 450 hectares of soybeans
<h3>How to solve algebra word problem?</h3>
He grows corn, wheat and soya beans on the farm of 1200 hectares. Thus;
C + W + S = 12 ----(1)
It costs $45 per hectare to grow corn, $60 to grow wheat, and $50 to grow soybeans. Thus;
45C + 60W + 50S = 63750 -----(2)
He will grow twice as many hectares of wheat as corn. Thus;
W = 2C ------(3)
Put 2C for W in eq 1 and eq 2 to get;
C + 2C + S = 1200
3C + S = 1200 -----(4)
45C + 60(2C) + 50S = 63750
45C + 120C + 50S = 63750
165C + 50S = 63750 ------(5)
Solving eq 4 and 5 simultaneosly gives;
C = 250 and W = 500
Thus; S = 1200 - 3(250)
S = 450
Read more about algebra word problems at; brainly.com/question/13818690
Answer:
Among the different types of excavation protection system, as a way of preventing accidents against cave-ins, the sloping involves cutting back the trench wall at an angle inclined away from the excavation. Shoring requires installing aluminum hydraulic or other types of supports to prevent soil movement and cave-ins. Shielding protects workers by using trench boxes or other types of supports to prevent soil cave-ins (OSHA). In addition, the regulations do not allow employees to work on excavations where there is an accumulation of water. If this occurs, water on the site must be constantly removed by suitable equipment preventing water from accumulating. The entry of surface water into the excavations must also be prevented by means of diversion ditches, dam, or other suitable means.
Explanation:
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:
It is necessary to get the Reynold's number:
Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:
The overall heat transfer coefficient:
Here
Substituting values:
Answer: (a) 36.18mm
(b) 23.52
Explanation: see attachment
