1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
alexgriva [62]
3 years ago
12

Question 11 (1 point)

Engineering
1 answer:
kirill115 [55]3 years ago
4 0

Answer:

  False

Explanation:

Bella counts products in finished goods inventory and she counts kits in various stages of manufacturing.

You might be interested in
1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil
kakasveta [241]

Answer:

a) n = 1000\,users, b)\Delta t_{min} = \frac{1}{30}\,h, \Delta t_{max} = \frac{\sqrt{2} }{30}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h, c) n = 10000000\,users, \Delta t_{min} = \frac{1}{3000}\,h, \Delta t_{max} = \frac{\sqrt{2} }{3000}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

Explanation:

a) The total number of users that can be accomodated in the system is:

n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )

n = 1000\,users

b) The length of the side of each cell is:

l = \sqrt{1\,km^{2}}

l = 1\,km

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{30}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{30}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h

c) The total number of users that can be accomodated in the system is:

n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )

n = 10000000\,users

The length of each side of the cell is:

l = \sqrt{100\,m^{2}}

l = 10\,m

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{3000}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

8 0
3 years ago
Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.
Rufina [12.5K]

Answer:

a) \dot m = 16.168\,\frac{kg}{s}, b) v_{out} = 680.590\,\frac{m}{s}, c) \dot W_{out} = 18276.307\,kW

Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0

Energy Balance

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

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

\nu_{in} = 0.055665\,\frac{m^{3}}{kg}

h_{in} = 3650.6\,\frac{kJ}{kg}

Outlet (Liquid-Vapor Mix)

\nu_{out} = 5.89328\,\frac{m^{3}}{kg}

h_{out} = 2500.2\,\frac{kJ}{kg}

a) The mass flow rate of the steam is:

\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}

\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }

\dot m = 16.168\,\frac{kg}{s}

b) The exit velocity of steam is:

\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}

v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}

v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}

v_{out} = 680.590\,\frac{m}{s}

c) The power output of the steam turbine is:

\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})

\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)

\dot W_{out} = 18276.307\,kW

6 0
3 years ago
When checking for a no-star concern, you notice that an engine has no spark Technician A says to turn on the ignition engine (en
lbvjy [14]

Answer:

Technician B

Explanation:

Technician B is correct in his argument. This is because according to what he said, as the computer pulses stimuli the coil will turn on and off, promoting an increase in the voltage that will cause the fluctuation. Technician A is incorrect because the procedure he indicated imposes that the voltage is checked at the negative terminal and not at the positive.

5 0
2 years ago
Integer to Float Conversion All labs must be done during lab time. Each labs worth 10 points The lab can be hand in next day wit
andrew-mc [135]

Answer:

Code explained below

Explanation:

.data

msg1: .asciiz "Please input a temperature in celsius: "

msg2: .asciiz "The temperature in Fahrenheit is: => "

num: .float 0.0

.text

main:

#print the msg1

li $v0, 4

la $a0, msg1

syscall

#read the float value from user

li $v0,6 #read float syscall value is $v0

syscall #read value stored in $f0

#formula for celsius to fahrenheit is

#(temperature(C)* 9/5)+32

#li.s means load immediate float

#copy value 9.0 to $f2

li.s $f2,9.0  

#copy value 5.0 to $f3

li.s $f3,5.0

# following instructions performs: 9/5

#div.s - division of two float numbers

#divide $f2 and f3.Result will stores in $f1

div.s $f1,$f2,$f3

#following instruction performs: temperature(C) * (9/5)

#multiple $f1 and $f0.Result stored in $f1

mul.s $f1,$f1,$f0

#copy value 32 to $f4

li.s $f4,32.0

#following instruction performs: (temperature(C) * (9/5))+32

#add $f1 and $f4.Result stores in $f1

add.s $f1,$f1,$f4

#store float from $f1 to num

s.s $f1,num

#print the msg2

li $v0, 4 #print string syscall value is 4

la $a0, msg2 #copy address of msg2 to $a0

#print the float

syscall

li $v0,2 #print float syscall value is 2

l.s $f12,num #load value in num to $f12

syscall

#terminate the program

li $v0, 10 #terminate the program syscall value is 10

syscall

4 0
3 years ago
A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte
Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

a(t)=5(1-\frac{t}{15})

a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})

Integrating both sides we get

\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m

Thus the remaining 125 meters will be covered with a constant speed of

v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s

in time equalling t_{2}=\frac{125}{37.5}=3.33seconds

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0
3 years ago
Other questions:
  • Consider a cubical furnace with a side length of 3 m. The top surface is maintained at 700 K. The base surface has emissivity of
    13·1 answer
  • A student is using a 12.9 ft ramp to raise an object 6 ft above the ground.
    5·1 answer
  • A bus travels the 100 miles between A and B at 50 mi/h and then another 100 miles between B and C at 70 mi/h.
    6·1 answer
  • A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as
    5·1 answer
  • Everyone has only one learning style. True or false? hurry pleasle this exp carees class
    11·1 answer
  • A misfire code is a type ____ DTC<br> A) 1 or 2<br> B) a or b<br> C) c or d<br> D l or ll
    15·1 answer
  • La iluminación de la superficie de un patio amplio es 1600 lx cuando el ángulo de elevación del sol 53°. Calcular la iluminación
    15·1 answer
  • For the floor plan shown, if a = sm b= 8m, specify type of Load on Beam AHS<br> D<br> B В
    10·1 answer
  • What is the difference between the elements of design and the principles of design? Define at
    7·1 answer
  • Fast plz-The mirror check may involve ______________.
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!