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Colt1911 [192]
2 years ago
8

A master stud pattern is laid out somewhat like a?

Engineering
1 answer:
Svetradugi [14.3K]2 years ago
3 0

Answer:

••• like a story pole but has information for only one portion of the wall. system. methods and materials of construction.

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What is the mechanical advantage of a simple block and tackle with one rope?
NISA [10]

Answer:

A block and tackle is characterized by the use of a single continuous rope to transmit a tension force around one or more pulleys to lift or move a load. Its mechanical advantage is the number of parts of the rope that act on the load. The mechanical advantage of a tackle dictates how much easier it is to haul or lift the load.

Explanation:

Hope this helps- have a good day ^w

6 0
3 years ago
Read 2 more answers
Which of the following explains the purpose of a convertible screwdriver?
aliya0001 [1]

Answer:

The correct option is;

To be able to switch between different heads quickly and easily

Explanation:

A screwdriver is a tool or device that has a grip handle which surrounds an extension shaft that has a forged tip that fits into the complementary groove  on a screw head such that by turning the handle while the forged head of the screwdriver sits in a screw head, the screw is turned

A convertible screw driver is built with the capability to easily change the tip or other attributes of the screwdriver quickly.

7 0
3 years ago
Air exits a compressor operating at steady-state, steady-flow conditions at 150 oC, 825 kPa, with a velocity of 10 m/s through a
ioda

Answer:

a) Qe = 0.01963 m^3 / s , mass flow rate m^ = 0.1334 kg/s

b) Inlet cross sectional area = Ai = 0.11217 m^2 , Qi = 0.11217 m^3 / s    

Explanation:

Given:-

- The compressor exit conditions are given as follows:

                  Pressure ( Pe ) = 825 KPa

                  Temperature ( Te ) = 150°C

                  Velocity ( Ve ) = 10 m/s

                  Diameter ( de ) = 5.0 cm

Solution:-

- Define inlet parameters:

                  Pressure = Pi = 100 KPa

                  Temperature = Ti = 20.0

                  Velocity = Vi = 1.0 m/s

                  Area = Ai

- From definition the volumetric flow rate at outlet ( Qe ) is determined by the following equation:

                   Qe = Ae*Ve

Where,

           Ae: The exit cross sectional area

                   Ae = π*de^2 / 4

Therefore,

                  Qe = Ve*π*de^2 / 4

                  Qe = 10*π*0.05^2 / 4

                  Qe = 0.01963 m^3 / s

 

- To determine the mass flow rate ( m^ ) through the compressor we need to determine the density of air at exit using exit conditions.

- We will assume air to be an ideal gas. Thus using the ideal gas state equation we have:

                   Pe / ρe = R*Te  

Where,

           Te: The absolute temperature at exit

           ρe: The density of air at exit

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρe = Pe / (R*Te)

                ρe = 825 / (0.287*( 273 + 150 ) )

                ρe = 6.79566 kg/m^3

- The mass flow rate ( m^ ) is given:

               m^ = ρe*Qe

                     = ( 6.79566 )*( 0.01963 )

                     = 0.1334 kg/s

- We will use the "continuity equation " for steady state flow inside the compressor i.e mass flow rate remains constant:

              m^ = ρe*Ae*Ve = ρi*Ai*Vi

- Density of air at inlet using inlet conditions. Again, using the ideal gas state equation:

               Pi / ρi = R*Ti  

Where,

           Ti: The absolute temperature at inlet

           ρi: The density of air at inlet

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρi = Pi / (R*Ti)

                ρi = 100 / (0.287*( 273 + 20 ) )

                ρi = 1.18918 kg/m^3

Using continuity expression:

               Ai = m^ / ρi*Vi

               Ai = 0.1334 / 1.18918*1

               Ai = 0.11217 m^2          

- From definition the volumetric flow rate at inlet ( Qi ) is determined by the following equation:

                   Qi = Ai*Vi

Where,

           Ai: The inlet cross sectional area

                  Qi = 0.11217*1

                  Qi = 0.11217 m^3 / s    

- The equations that will help us with required plots are:

Inlet cross section area ( Ai )

                Ai = m^ / ρi*Vi  

                Ai = 0.1334 / 1.18918*Vi

                Ai ( V ) = 0.11217 / Vi   .... Eq 1

Inlet flow rate ( Qi ):

                Qi = 0.11217 m^3 / s ... constant  Eq 2

               

6 0
3 years ago
9.For a single-frequency sine wave modulating signal of 3 kHz with a carrier frequency of 36 MHz, what is the spacing between si
nikdorinn [45]

The spacing between sidebands is equal to 6 kHz.

<u>Given the following data:</u>

  • Modulating signal = 3 kHz.
  • Carrier frequency = 36 MHz.

<h3>What is a sideband?</h3>

A sideband can be defined as a band of frequencies that are lower or higher than the carrier frequency due to the modulation process. Thus, it will either be lower than or higher than the carrier frequency.

Generally, the frequency of the modulating signal is equal to the spacing between the sidebands. Therefore, a modulating signal of 3 kHz simply means that the lower sideband is <u>3 kHz</u> higher while the upper sideband is <u>3 kHz</u> lower.

Spacing = 3 kHz + 3 kHz = 6 kHz.

Read more on frequency here: brainly.com/question/3841958

8 0
2 years ago
Suppose you must remove an average of 3.9×108J of thermal energy per day to keep your house cool during the summer. Part A If yo
noname [10]

Answer:

The required mechanical work is required to reduce each day by 1.05×10^8 Joules.

Explanation:

Coefficient of Performance (COP) = Q/W

Q is thermal energy absorbed by the air conditioner

W is mechanical work done

Q = 3.9×10^8 J

COP of old air conditioner = 2.3

W = Q/COP = 3.9×10^8/2.3 = 1.70×10^8 J

COP of new air conditioner = 6

W = Q/COP = 3.9×10^8/6 = 6.5×10^7 J

Reduction in mechanical work = (1.7×10^8) - (6.5×10^7) = 1.05×10^8 J

3 0
3 years ago
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