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myrzilka [38]
3 years ago
11

Among Us Map-Skeld Imposter 3 code- DBZPTQ

Engineering
2 answers:
Yuki888 [10]3 years ago
7 0

Answer:

0:  <em>NOICE  </em>:0

Explanation:

antoniya [11.8K]3 years ago
7 0
I’m surprised this game is still popping. Lol
You might be interested in
___________ is NOT a common injury that an automotive tech may experience at work.
Degger [83]

Answer:The most common injuries were sprains/strains (39% of the total), lacerations (22%), and contusions (15%). Forty-nine percent of the injuries resulted in one or more lost or restricted workdays; 25% resulted in 7 or more lost or restricted workdays.

Explanation:

The most common injuries were sprains/strains (39% of the total), lacerations (22%), and contusions (15%). Forty-nine percent of the injuries resulted in one or more lost or restricted workdays; 25% resulted in 7 or more lost or restricted workdays.

7 0
3 years ago
A cylindrical specimen of some metal alloy having an elastic modulus of 106 GPa and an original cross-sectional diameter of 3.9
kiruha [24]

Answer:

L= 312.75 mm

Explanation:

given data

elastic modulus E = 106 GPa

cross-sectional diameter d = 3.9 mm

tensile load F = 1660 N

maximum allowable elongation ΔL = 0.41 mm

to find out

maximum length of the specimen before deformation

solution

we will apply here allowable elongation equation that is express as

ΔL =     \dfrac{FL}{AE}     ....................1

put here value and we get L

L   =    \dfrac{0.41\times 10^{-3}\times \dfrac{\pi}{4}\times (3.9\times 10^{-3})^2\times 106\times 10^9}{1660}

solve it we get

L = 0.312752 m

L= 312.75 mm

8 0
4 years ago
A tensile test uses a test specimen that has a gage length of 50 mm and an area = 206 mm2. During the test, the specimen yields
vampirchik [111]

Answer:

The percent elongation in the length of the specimen is 42%

Explanation:

Given that:

The gage length of the original test specimen  L_o = 50 mm

The final gage length L_f = 71 mm

The area = 206 mm²

maximum load  =  162,699 N

To determine the percent elongation in %, we use the formula:

\%EL = \dfrac{L_f-L_o}{L_o}\times 100

\%EL = \dfrac{71 \ mm-50 \ mm}{50 \ mm}\times 100

\%EL = \dfrac{21 mm}{50 \ mm}\times 100

\%EL = 0.42 \times 100

\mathbf{\%EL = 42 \%}

The percent elongation in the length of the specimen is 42%

4 0
4 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
4 years ago
Calculate the amount of power (in Watts) required to move an object weighing 762 N from point A to point B within 29 seconds. Di
Tresset [83]

Answer:

0.556 Watts

Explanation:

w = Weight of object = 762 N

s = Distance = 5 m

t = Time taken = 29 seconds

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

s=ut+\frac{1}{2}at^2\\\Rightarrow a=\frac{2\times (s-ut)}{t^2}\\\Rightarrow a=\frac{2\times (5-0)}{29^2}=\frac{10}{481}

Mass of the body

m=\frac{w}{g}=\frac{762}{9.81}

Force required to move the body

F=ma\\\Righarrow F=\frac{762}{9.81}\times \frac{10}{481}

Velocity of object

v=u+at\\\Rightarrow v=0+\frac{10}{481}\times 29\\\Rightarrow v=\frac{10}{29}

Power

P=Fv\\\Rightarrow P=\frac{762}{9.81}\times \frac{10}{481}\times \frac{10}{29}=0.556\ W

∴ Amount of power required to move the object is 0.556 Watts

7 0
4 years ago
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