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garik1379 [7]
4 years ago
6

For a statically indeterminate axially loaded member, Group of answer choices The total deflection between end A and end B of an

axially loaded member must not be zero. The summation of the forces is not zero. The summation of the reaction forces is equal to the applied load. The applied load must exceed the total reaction forces. The compatibility condition cannot be satisfied, since it is indeterminate.
Engineering
1 answer:
blsea [12.9K]4 years ago
4 0

Answer: The summation of the reaction forces is equal to the applied load.

Explanation:

When solving an indeterminate structure, it is important for one to satisfy the force-displacement requirements, compatibility, and the equilibrium of the structure.

With regards to the above question, for a statically indeterminate axially loaded member, the summation of the reaction forces will be equal to applied load. Here, the summation of the reactive forces will then be zero. This is a condition that is necessary to solve the unknown reaction forces.

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A ten-station transfer machine has an ideal cycle time of 30 sec. The frequency of line stops is 0.075 stops per cycle. When a l
goldfiish [28.3K]

Answer:

62.5%

Explanation:

We are given that

A ten-station transfer machine has ideal cycle time=30 sec=\frac{30}{60}=0.5 min

Frequency of line=0.075 stops per cycle

Average time=4 min

We have to determine the line efficiency

T_p=0.5+0.075(4)=0.5+0.3=0.8

Line efficiency=\frac{0.5}{0.8}\times 100=62.5%

Hence, the line efficiency=62.5%

8 0
4 years ago
The explosion of a hydrogen bomb can be approximated by a fireball with a temperature of 7200 K, according to a report published
Iteru [2.4K]

Answer:

a

The rate of radiation of the energy is  E_r = 1.523747635*10^9 W/m^2

b

The irradiation is  G =46.177\ kW/m^2

c

The amount of energy absorbed is E_B = 461.772 KJ

d

The oak Tree would catch fire because the temperature of the blast(7200 K) is higher than the flammability limit (650 K) of the oak tree and secondly the thickness is very small

Explanation:

  From the question we are told that

        The  temperature is  T =  7200K

        The diameter of the ball is  d = 1.5 km = 1.5 *1000 = 1500m

       Hence the radius  == \frac{1500}{2} = 750m

 The total energy radiated can be mathematically represented as

                         E = \sigma A T^4

Where \sigma is the Stefan-Boltzmann constant \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5.67*10^{-8} W \ \cdot m^{-2} K^{-4}

            A is the area of a sphere  = \pi d^2  = 3.142 * 1500^2 = 7.069500 *10 ^6\ m^2

 Substituting values we have

                    E = 5,67*10^{-8} * 7.069500*10^6 * 7200^4

                        =1.077*10^{15} W

Now the state of the energy is mathematically represented as

                           Rate  \ of \ energy \ radiation (E_r)= \frac{E}{A} = \sigma T^4

                                                            = 5.67*10^{-8} * 7200^2

                                                            = 1.523747635*10^9 W/m^2

A sketch illustrating the b part of the question is shown on the first uploaded image

     looking at the height at which the blast occurs(16km) as compared to the height of the wall we notice that the height of the wall is negligibly small

      from the diagram x can be calculated as follows

                      x = \sqrt{40^2 + 16^2}

                        = 43.0813 Km

This value of x represents the radius of the blast(assuming it is spherical ) when it is at that wall

Now the irradiation G is mathematically represented as

                              G = \frac{E}{4 \pi r^2}

Here r = 43.0813 Km = 43.0813 × 1000 = 43081.3 m

                            G= \frac{1.077*10^15}{4 \pi (431081.3^2)}

                                G =46.177\ kW/m^2

Generally the amount of energy absorbed can be mathematically represented as

                            Amount \ of \ energy \ absorbed \ (E_B ) = G * t

Where t is the time taken

       Therefore     E_B = 46.177 *10 = 461.77 KJ

       

                         

                       

             

   

6 0
3 years ago
Which of the following is the LEAST-Likely cause of tire wear? A) Underinflation B) Braking C) Acceleration D) Tire rotation
mihalych1998 [28]

Answer:

Tire rotation is the least likely cause of tire wear. So, the option D is correct.

Explanation:

Step1

Under-inflation is the process of tire failure under low pressure. This contributes the wear on tire.

Step2

On breaking, kinetic energy changes to heat energy because of rubbing of tire. So, rubbing action increases the wear on the tire.

Step3

Acceleration on the vehicle increases the rubbing action as well as the wear and tear on the tire. So, acceleration is an also a major cause of tire wear.

Step4

Tire rotation has least amount of wear and tear due to no rubbing action.  It has less amount surface contact with the surface in rotation.  

Thus, tire rotation is the least likely cause of tire wear. So, the option D is correct.  

4 0
3 years ago
Which equation can be used to find x, the length of the hypotenuse of the right triangle? A triangle has side lengths 63, 16, x.
Kryger [21]

Answer: 16 squared + 63 squared = x squared

Explanation:

Hi, since we have a right triangle we have to apply the Pythagorean Theorem:  

c^2 = a^2 + b^2  

Where c is the hypotenuse of the triangle (the longest side) and a and b are the other sides.  

Replacing with the values given:  

x^2 = 63^2 + 16^2  

So, the correct option is  

16 squared + 63 squared = x squared

Feel free to ask for more if needed or if you did not understand something.  

3 0
4 years ago
Read 2 more answers
Electronic highway message boards communicate how
kakasveta [241]

Answer:

with words giving information about the road ahead.

Explanation:

7 0
3 years ago
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