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I am Lyosha [343]
3 years ago
13

A European car manufacturer reports that the fuel efficiency of the new MicroCar is 48.5 km/L highway and 42.0 km/L city. What a

re the equivalent fuel efficiency rates in miles per gallon?
Engineering
1 answer:
statuscvo [17]3 years ago
3 0

Answer:

Fuel efficiency for highway = 114.08 miles/gallon

Fuel efficiency for city = 98.79 miles/gallon

Explanation:

1 gallon = 3.7854 litres

1 mile = 1.6093 km

Let's first convert the efficiency to km/gallon:

48.5 km/litre = (48.5 * 3.7854) km/gallon

48.5 km/litre =  183.5919 km/gallon (highway)

42.0 km/litre = (42.0 * 3.7854) km/gallon

42.0 km/litre = 158.9868 km/gallon (city)

Next, we convert these to miles/gallon:

183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon

183.5919 km/gallon = 114.08 miles/gallon (highway)

158.9868 km/gallon = (158.9868 /1.6093) miles/gallon

158.9868 km/gallon = 98.79 miles/gallon (city)

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Water is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 12 MPa, and the condenser pressure i
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Answer:

\dot Q_{in} = 372.239\,MW

Explanation:

The water enters to the pump as saturated liquid and equation is modelled after the First Law of Thermodynamics:

w_{in} + h_{in}- h_{out} = 0

h_{out} = w_{in}+h_{in}

h_{out} = 12\,\frac{kJ}{kg} + 191.81\,\frac{kJ}{kg}

h_{out} = 203.81\,\frac{kJ}{kg}

The boiler heats the water to the state of saturated vapor, whose specific enthalpy is:

h_{out} = 2685.4\,\frac{kJ}{kg}

The rate of heat transfer in the boiler is:

\dot Q_{in} = \left(150\,\frac{kg}{s}\right)\cdot \left(2685.4\,\frac{kJ}{kg}-203.81\,\frac{kJ}{kg} \right)\cdot \left(\frac{1\,MW}{1000\,kW} \right)

\dot Q_{in} = 372.239\,MW

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3 years ago
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Answer:

yes

Explanation:

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Please guys do you need technical drawing to be a robotics or software engineer?
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Explanation:

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This elementary problem begins to explore propagation delayand transmission delay, two central concepts in data networking. Cons
telo118 [61]

Explanation:

(a)

Here, distance between hosts A and B is m meters and, propagation speed along the link is s meter/sec

Hence, propagation delay, d_{prop} = m/sec (s)

(b)

Here, size of the packet is L bits

And the transmission rate of the link is R bps

Hence, the transmission time of the packet,  d_{trans} = L/R

(c)

As we know, end-to-end delay or total no delay,

\mathrm{d}_{\text {nodal }}=\mathrm{d}_{\text {proc }}+\mathrm{d}_{\text {quar }}+d_{\max }+d_{\text {prop }}

Here,  $\mathrm{d}_{\text {rroc }}$ and $\mathrm{d}_{\text {quat }}$ \\Hence, $\mathrm{d}_{\text {rodal }}=\mathrm{d}_{\text {trass }}+\mathrm{d}_{\text {prop }}$ \\We know, $\mathrm{d}_{\text {trax }}=\mathrm{L} / \mathrm{R}$ sec and $\mathrm{d}_{\text {vapp }}=\mathrm{m} / \mathrm{s}$ sec\text { Hence, } {d_{\text {nodal }}}=\mathrm{L} / \mathrm{R}+\mathrm{m} / \mathrm{s} \text { seconds }

(d)

The expression, time time $t=d_{\text {trans }}$ means the\at time since transmission started is equal to transmission delay.

As we know, transmission delay is the time taken by host to push out the packet.

Hence, at time $t=d_{\text {trans }}$ the last bit of the packet has been pushed out or transmitted.

(e)

If \ d_{prop} >d_{trans}

Then, at time $t=d_{\text {trans }}$ the bit has been transmitted from host A, but to condition (1),  the first bit has not reached B.

(f)

If \ d_{prop}

Then, at time $t=d_{\text {trans }}$, the first bit has reached destination on B

Here,s=2.5 \times 10^{8} \mathrm{sec}

\begin{aligned}&\mathrm{L}=100 \mathrm{Bits} \text { and }\\&\mathrm{R}=28 \mathrm{kbps} \text { or } 28 \times 1000 \mathrm{bps}\end{aligned}

It's given that \ d_{prop} =d_{trans}

Hence,

        \begin{aligned}\ & \frac{L}{R}=\frac{m}{s} \\m &=s \frac{L}{R} \\&=\frac{2.5 \times 10^{8} \times 100}{28 \times 1000} \\&=892.9 \mathrm{km}\end{aligned}

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3 years ago
The ABC Corporation manufactures and sells two products: T1 and T2. 20XX budget for the company is given below:
Sliva [168]

Answer:

The ABC Corporation

a) Total Expected Revenue (in dollars) for 20XX:

Revenue from T1 = 60,000 x $165 = $26,400,000

Revenue from T2 = 40,000 x $250 = $10,000,000

Total Revenue from T1 and T2 = $36,400,000

b) Production Level (in units) for T1 and T2

                                           T1                       T2

Total Units sold             160,000           40,000

Add Closing Inventory   25,000             9,000

Units Available for sale 185,000           49,000

less opening inventory  20,000             8,000

Production Level          165,000 units 41,000 units

c) Total Direct Material Purchases (in dollars):

Cost of direct materials used    T1                T2

A:       (165,000 x 4 x $12)   $7,920,000   $2,460,000 (41,000 x 5 x $12)

B:       (165,000 x 2 x $5)       1,650,000          615,000 (41,000 x 3 x $5)

C:                                                           0          123,000 (41,000 x 1 x$3)

Total cost                            $9,570,000     $3,198,000 Total = $12,768,000

Cost of direct per unit = $58 ($9,570,000/165,000) for T1 and $78 ($3,198,000/41,000) for T2

Cost of direct materials used for production $12,768,000

Cost of closing direct materials:

                 A  (36,000 x $12)  $432,000

                 B (32,000 x $5)        160,000

                 C (7,000 x $3)            21,000             $613,000

Cost of direct materials available for prodn   $13,381,000

Less cost of beginning direct materials:

                 A  (32,000 x $12)        $384,000

                 B  (29,000 x $5)            145,000

                 C  (6,000 x $3)                18,000        $547,000

Cost of direct materials purchases               $12,834,000

d) The Total Direct Manufacturing Labor Cost (in dollars):

                                             T1                         T2

Direct labor per unit              2 hours                  3 hours

Direct labor rate per hour    $12                        $16

Direct labor cost per unit   $24                          $48

Production level              165,000 units        41,000 units

Labor Cost ($)                $3,960,000        $1,968,000

Total labor cost  $5,928,000 ($3,960,000 + $1,968,000)

e) Total Overhead cost (in dollars):

Overhead rate  = $20 per labor hour

Overhead cost per unit: T1 = $40 ($20 x 2) and T2 = $60 ($20 x 3)

T1 overhead = $20 x 2  x 165,000) = $6,600,000

T2 overhead = $20 x 3 x 41,000) =    $2,460,000

Total Overhead cost =                        $9,060,000

Cost of goods produced:

Cost of opening inventory of materials  = $547,000

Purchases of directials materials             12,834,000

less closing inventory of materials     =      $613,000

Cost of materials used for production    12,768,000

add Labor cost                                           5,928,000

add Overhead cost                                    9,060,000

Total production cost                            $27,756,000

f) Total cost of goods sold (in dollars):

Cost of opening inventory =          $3,928,000

Total Production cost             =    $27,756,000

Cost of goods available for sale  $31,684,000

Less cost of closing inventory       $4,724,000

Total cost of goods sold            $26,960,000

g) Total expected operating income (in dollars)

Sales Revenue:  T1 and T2  $36,400,000

Cost of goods sold                 26,960,000

Gross profit                             $9,440,000

less marketing & distribution      400,000

Total Expected Operating Income = $9,040,000

Explanation:

a) Cost of beginning inventory of finished goods:

T1, (Direct materials + Labor + Overhead) X inventory units =

T1 = 20,000 x ($58 + 24 + 40) = $2,440,000

T2 = 8,000 ($78 + 48 + 60) = $1,488,000

Total cost of beginning inventory = $3,928,000

b) Cost of closing Inventory of finished goods:

T1 = 25,000 x ($58 + 24 + 40) = $3,050,000

T2 = 9,000 ($78 + 48 + 60) = $1,674,000

Total cost of closing inventory = $4,724,000

5 0
3 years ago
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