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Oxana [17]
2 years ago
12

To make 1000 containers of ice cream you need: 600 gallons of milk, 275 gallons of cream, and 120 gallons of flavor. Each ingred

ient is contained in tanks like the ones pictured below. Note the capacity of each container. Adjustments to one ingredient must be followed by corresponding adjustments to the other ingredients. If you decrease the amount of milk by 10%, how many gallons must the cream and flavor be decreased?
Engineering
1 answer:
Kamila [148]2 years ago
3 0

Answer:

For the cream, 32 gallons should be reduced and 12 gallons should be decreased for flavor.

Explanation:

To prepare a total of 1000 gallons of ice cream you need 600 gallons of milk, 275 gallons of cream and 120 gallons of flavor, therefore we must calculate the percentages of each ingredient, as follows:

%milk=(600/1000)x100=60%

%cream=(275/1000)x100=27.5%

%flavor=(120/1000)x100=12%

If you reduce the amount of milk by 10% you have:

Milk quantity=600 gallons-(600x0.1)=540 gallons

To maintain the same percentages of each ingredient, you must make a rule of three to know the amount of cream and flavor that would need to be used with the 540 gallons of milk. The rule of three is as follows:

540 gallons of milk------------------60%

x gallons of cream--------------------27.5%

Clearing the x:

x gallons of cream=(540x27.5)/60=243 gallons

In the same way for flavor:

540 gallons of milk------------------60%

x gallons of flavor--------------------12%

Clearing the x:

x gallons of flavor=(540x12)/60=108 gallons

Verifying that they meet the percentages that were calculated before:

Total amount of ice cream=540+243+108=891 gallons

Calculate the percentages of each ingredient:

%milk=(540/891)x100=60.6%

%cream=(243/891)x100=27.3%

%flavor=(108/891)x100=12.1%

As can be seen, it is found that approximately the same percentages calculated above are met. Therefore, we can already calculate the amount by which the cream should be reduced and the flavor.

For the cream:

Gallons of cream=275-243=32 gallons

For the flavor:

Gallons of flavor=120-108=12 gallons

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A steam power plant operates on an ideal reheat- regenerative Rankine cycle and has a net power output of 80 MW. Steam enters th
trasher [3.6K]

Answer:

flow(m) = 54.45 kg/s

thermal efficiency u = 44.48%

Explanation:

Given:

- P_1 = P_8 = 10 KPa

- P_2 = P_3 = P_6 = P_7 = 800 KPa

- P_4 = P_5 = 10,000 KPa

- T_5 = 550 C

- T_7 = 500 C

- Power Output P = 80 MW

Find:

-  The mass flow rate of steam through the boiler

-  The thermal efficiency of the cycle.

Solution:

State 1:

P_1 = 10 KPa , saturated liquid

h_1 = 192 KJ/kg

v_1 = 0.00101 m^3 / kg

State 2:

P_2 = 800 KPa , constant volume process work done:

h_2 = h_1 + v_1 * ( P_2 - P_1)

h_2 = 192 + 0.00101*(790) = 192.80 KJ/kg

State 3:

P_3 = 800 KPa , saturated liquid

h_3 = 721 KJ/kg

v_3 = 0.00111 m^3 / kg

State 4:

P_4 = 10,000 KPa , constant volume process work done:

h_4 = h_3 + v_3 * ( P_4 - P_3)

h_4 = 721 + 0.00111*(9200) = 731.21 KJ/kg

State 5:

P_5 = 10,000 KPa , T_5 = 550 C

h_5 = 3500 KJ/kg

s_5 = 6.760 KJ/kgK

State 6:

P_6 = 800 KPa , s_5 = s_6 = 6.760 KJ/kgK

h_6 = 2810 KJ/kg

State 7:

P_7 = 800 KPa , T_7 = 500 C

h_7 = 3480 KJ/kg

s_7 = 7.870 KJ/kgK

State 8:

P_8 = 10 KPa , s_8 = s_7 = 7.870 KJ/kgK

h_8 = 2490 KJ/kg

- Fraction of steam y = flow(m_6 / m_3).

- Use energy balance of steam bleed and cold feed-water:

                                        E_6 + E_2 = E_3

               flow(m_6)*h_6 + flow(m_2)*h_3 = flow(m_3)*h_3

                                    y*h_6 + (1-y)*h_3 = h_3

                                  y*2810 + (1-y)*192.8 = 721

Compute y:                          y = 0.2018

- Heat produced by the boiler q_b:

                             q_b = h_5 - h_4 +(1-y)*(h_7 - h_8)

                    q_b = 3500 -731.21 + ( 1 - 0.2018)*(3480 - 2810)

Compute q_b:               q_b = 3303.58 KJ/ kg

-Heat dissipated by the condenser q_c:

                                       q_c = (1-y)*(h_8 - h_1)

                                 q_c= ( 1 + 0.2018)*(2810 - 192)

Compute q_c:               q_c = 1834.26 KJ/ kg

- Net power output w_net:

                                     w_net = q_b - q_c

                                w_net = 3303.58 - 1834.26

                                    w_net = 1469.32 KJ/kg

- Given out put P = 80,000 KW

                                     flow(m) = P / w_net

compute flow(m)          flow(m) = 80,000 /1469.32 = 54.45 kg/s

- Thermal efficiency u:

                                     u = 1 - (q_c / q_b)

                                     u = 1 - (1834.26/3303.58)

                                     u = 44.48 %

5 0
2 years ago
A steady tensile load of 5.00kN is applied to a square bar, 12mm on a side and having a length of 1.65m. compute the stress in t
Shtirlitz [24]

Answer:

The stress in the bar is 34.72 MPa.

The design factor (DF) for each case is:

A) DF=0.17

B) DF=0.09

C) DF=0.125

D) DF=0.12

E) DF=0.039

F) DF=1.26

G) DF=5.5

Explanation:

The design factor is the relation between design stress and failure stress. In the case of ductile materials like metals, the failure stress considered is the yield stress. In the case of plastics or ceramics, the failure stress considered is the breaking stress (ultimate stress). If the design factor is less than 1, the structure or bar will endure the applied stress. By the opposite side, when the DF is higher than 1, the structure will collapse or the bar will break.

we will calculate the design stress in this case:

\displaystyle \sigma_{dis}=\frac{T_l}{Sup}=\frac{5.00KN}{(12\cdot10^{-3}m)^2}=34.72MPa

The design factor for metals is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{y}}

The design factor for plastic and ceramics is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{u}}

We now need to know the yield stress or the ultimate stress for each material. We use the AISI and ASTM charts for steels, materials charts for non-ferrous materials and plastics safety charts for the plastic materials.

For these cases:

A) The yield stress of AISI 120 hot-rolled steel (actually is AISI 1020) is 205 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{205MPa}=0.17

B) The yield stress of AISI 8650 OQT 1000 steel is 385 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{385MPa}=0.09

C) The yield stress of ductile iron A536-84 (60-40-18) is 40Kpsi, this is 275.8 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{275.8MPa}=0.125

D) The yield stress of aluminum allot 6061-T6 is 290 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{290MPa}=0.12

E) The yield stress of titanium alloy Ti-6Al-4V annealed (certified by manufacturers) is 880 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{880MPa}=0.039

F) The ultimate stress of rigid PVC plastic (certified by PVC Pipe Association) is 4Kpsi or 27.58 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{27.58 MPa}=1.26

In this case, the bar will break.

F) You have to consider that phenolic plastics are used as matrix in composite materials and seldom are used alone with no reinforcement. In this question is not explained if this material is reinforced or not, therefore I will use the ultimate stress of most pure phenolic plastics, in this case, 6.31 MPa:

DF=\displaystyle\frac{34.72MPa}{6.31 MPa}=5.5

This material will break.

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An optical mouse originally cost $31.85. Before it was removed from the store, it underwent the following changes in price. 27%
saul85 [17]

Answer:

b.

$43.21

Explanation:

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Your roommate wants to learn about organizational chart. You can tell her that these charts can tell you all of these about an o
mars1129 [50]

Answer:

Control mechanisms

Explanation:

Organizational chart of any company will give details of different aspects of the company such as the major sub-units of the organization with the names of team leaders for different sub-units, it can also give you the span of control and the division of work within the company. However, the chart can't show you control mechanisms of different departments.

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49. A solenoid coil with a resistance of 30 ohms and an inductance of 200 milli-henrys, is connected to a 230VAC, 50Hz supply. D
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Answer:

69.59 ohms

Explanation:

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L=200\ mH

F=50\ hZ

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X_L=2\times \pi\times f \times LX_L\\=2\times \pi \times 50\times 200\times 10^{-3}\\=62.8\ ohm

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