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Blababa [14]
3 years ago
6

Which 1 of the following did women NOT do during WWII?

Engineering
2 answers:
maria [59]3 years ago
4 0
What are the things we’re choosing from lma0
Anastaziya [24]3 years ago
3 0

Answer:

Didn't do what, there are no women as answers. Contact me and give me the women.

Explanation:

You might be interested in
What are the rigging devices used to move loads such as steel plates and sheet piles without the use of slings, but grip the loa
scoundrel [369]

The rigging device which are used to move loads without the use of slings, but grip the load by biting down and using jaw tension to secure the load, is lifting clamps.

<h3>What are the rigging devices?</h3>

The rigging devices are used to lift the objects and items when the safety is required. This device is used in the industries.

Types of rigging devices

  • Rigging hooks-These rigging device is used when the heavy load need to be lift.
  • Lifting clamps-Lifting clamp are used to lift the device with jaw tension to secure the load. In this, there is no use of slings.
  • Pulley and blocks-In the load is lifts with the help of block and pulley arrangement. This is a widely used rigging device.

Thus, the rigging device which are used to move loads without the use of slings, but grip the load by biting down and using jaw tension to secure the load, is lifting clamps.

Learn more about the rigging devices here;

brainly.com/question/8430576

#SPJ1

4 0
2 years ago
The two shafts of a Hooke’s coupling have their axes inclined at 20°.The shaft A revolves at a uniform speed of 1000 rpm. The sh
lapo4ka [179]

Answer:

33.429 N-m

Explanation:

Given :

Inclination angle of two shaft, α = 20°

Speed of shaft A, N_{A} = 1000 rpm

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

                                   = 0.1 m

Now we know that for maximum velocity,

\frac{N_{B}}{N_{A}} = \frac{cos\alpha }{1 - sin^{2}\alpha }

\frac{N_{B}}{1000} = \frac{cos20}{1 - sin^{2}20 }

N_{B} = 1064.1 rpm

Now we know

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

                                   = 0.1 m

Therefore moment of inertia of flywheel, I = m.k^{2}

                                                                      =30 X 0.1^{2}

                                                                     = 0.3 kg-m^{2}

Now torque on the output shaft

T₂ = I x ω

    = 0.3 X 1064.2 rpm

    = 0.3\times \frac{2\pi \times 1064.1}{60}

     = 33.429 N-m

Torque on the Shaft B is 33.429 N-m

4 0
3 years ago
The voltage across a device and the current through it are:
butalik [34]

Answer:

attached below

Explanation:

3 0
3 years ago
Read 2 more answers
What does it mean to say that PEER is a data-driven, consumer-centric, and comprehensive system?
Reika [66]

Answer:

have you heard of gnoogle?

Explanation:have you heard of goongle?

3 0
3 years ago
Read 2 more answers
Someone has suggested that the air-standard Otto cycle is more accurate if the two polytropic processes are replaced with isentr
omeli [17]

Answer:

q_net,in = 585.8 KJ/kg

q_net,out = 304 KJ/kg

n = 0.481

Explanation:

Given:

- The compression ratio r = 8

- The pressure at state 1, P_1 = 95 KPa

- The minimum temperature at state 1, T_L = 15 C

- The maximum temperature T_H = 900 C

- Poly tropic index n = 1.3

Find:

a) Determine the heat transferred to and rejected from this cycle

b) cycle’s thermal efficiency

Solution:

- For process 1-2, heat is rejected to sink throughout. The Amount of heat rejected q_1,2, can be computed by performing a Energy balance as follows:

                                   W_out - Q_out = Δ u_1,2

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

                         c_v*(T_2 - T_L) = R*(T_2 - T_L)/n-1 - q_1,2

- Using polytropic relation we will convert T_2 = T_L*r^(n-1):

                  c_v*(T_L*r^(n-1) - T_L) = R*(T_1*r^(n-1) - T_L)/n-1 - q_1,2

- Hence, we have:

                             q_1,2 = T_L *(r^(n-1) - 1)* ( (R/n-1) - c_v)

- Plug in the values:

                             q_1,2 = 288 *(8^(1.3-1) - 1)* ( (0.287/1.3-1) - 0.718)

                            q_1,2= 60 KJ/kg

- For process 2-3, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

                                          Q_in = Δ u_2,3

                                         q_2,3 = u_3 - u_2

                                         q_2,3 = c_v*(T_H - T_2)  

- Again, using polytropic relation we will convert T_2 = T_L*r^(n-1):

                                         q_2,3 = c_v*(T_H - T_L*r^(n-1) )    

                                         q_2,3 = 0.718*(1173-288*8(1.3-1) )

                                        q_2,3 = 456 KJ/kg

- For process 3-4, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

                                     q_3,4 - w_in = Δ u_3,4

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

                           c_v*(T_4 - T_H) = - R*(T_4 - T_H)/1-n +  q_3,4

- Using polytropic relation we will convert T_4 = T_H*r^(1-n):

                  c_v*(T_H*r^(1-n) - T_H) = -R*(T_H*r^(1-n) - T_H)/n-1 + q_3,4

- Hence, we have:

                             q_3,4 = T_H *(r^(1-n) - 1)* ( (R/1-n) + c_v)

- Plug in the values:

                             q_3,4 = 1173 *(8^(1-1.3) - 1)* ( (0.287/1-1.3) - 0.718)

                            q_3,4= 129.8 KJ/kg

- For process 4-1, heat is lost from the system. The Amount of heat rejected q_4,1, can be computed by performing a Energy balance as follows:

                                          Q_out = Δ u_4,1

                                         q_4,1 = u_4 - u_1

                                         q_4,1 = c_v*(T_4 - T_L)  

- Again, using polytropic relation we will convert T_4 = T_H*r^(1-n):

                                         q_4,1 = c_v*(T_H*r^(1-n) - T_L )    

                                         q_4,1 = 0.718*(1173*8^(1-1.3) - 288 )

                                        q_4,1 = 244 KJ/kg

- The net gain in heat can be determined from process q_3,4 & q_2,3:

                                         q_net,in = q_3,4+q_2,3

                                         q_net,in = 129.8+456

                                         q_net,in = 585.8 KJ/kg

- The net loss of heat can be determined from process q_1,2 & q_4,1:

                                         q_net,out = q_4,1+q_1,2

                                         q_net,out = 244+60

                                         q_net,out = 304 KJ/kg

- The thermal Efficiency of a Otto Cycle can be calculated:

                                         n = 1 - q_net,out / q_net,in

                                         n = 1 - 304/585.8

                                         n = 0.481

6 0
3 years ago
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