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IgorLugansk [536]
2 years ago
5

Is reinforcement needed in a retaining wall

Engineering
2 answers:
velikii [3]2 years ago
8 0
Reinforcement means to rebuild. so yeah i guess you need to reinforce a walk
Charra [1.4K]2 years ago
5 0
Yea, bc you’ll need help reestablishing it
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Maggie discovered that a pipe in her basement has sprung a leak. She calls a plumber but in the meantime she grabs a roll of duc
dem82 [27]

Answer:

The answer is "Option a".

Explanation:

Myelination was the myelinization mechanism of a neuron axon. The endothelium is enveloped all around the axon and isolates the axon that inhibits the neuronal message from leaking with the other neuronal axons. Inside this example, therefore, its tubes tape worked similarly to those of myelin sheath, which stops brain transmission.

3 0
3 years ago
A machine used to lift motorcycles consists of an electric winch pulling on one supporting cable of a block and tackle system. T
Kobotan [32]

Answer: So you are dealing with maximum and minimum weights and you want to know what MINIMUM number of supporting strands for this block and tackle system are needed I believe. If so you are dealing with economic imbalances Though we are not worrying about money Right? Right we need physics which Physics study matter and how it moves You would need 8 STRANDS

Explanation: Step By Step

5 0
3 years ago
Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu
Sonja [21]

Answer:

a

The rate of work developed is \frac{\r W}{\r m}= 300kJ/kg

b

The rate of entropy produced within the turbine is   \frac{\sigma}{\r m}=  0.0861kJ/kg \cdot K

Explanation:

     From  the question we are told

          The rate at which heat is transferred is \frac{\r Q}{\r m } = -  30KJ/kg

the negative sign because the heat is transferred from the turbine

          The specific heat capacity of air is c_p = 1.1KJ/kg \cdot K

          The inlet temperature is  T_1 = 970K

          The outlet temperature is T_2 = 670K

           The pressure at the inlet of the turbine is p_1 = 400 kPa

          The pressure at the exist of the turbine is p_2 = 100kPa

           The temperature at outer surface is T_s = 315K

         The individual gas constant of air  R with a constant value R = 0.287kJ/kg \cdot K

The general equation for the turbine operating at steady state is \

               \r Q - \r W + \r m (h_1 - h_2) = 0

h is the enthalpy of the turbine and it is mathematically represented as          

        h = c_p T

The above equation becomes

             \r Q - \r W + \r m c_p(T_1 - T_2) = 0

              \frac{\r W}{\r m}  = \frac{\r Q}{\r m} + c_p (T_1 -T_2)

Where \r Q is the heat transfer from the turbine

           \r W is the work output from the turbine

            \r m is the mass flow rate of air

             \frac{\r W}{\r m} is the rate of work developed

Substituting values

              \frac{\r W}{\r m} =  (-30)+1.1(970-670)

                   \frac{\r W}{\r m}= 300kJ/kg

The general balance  equation for an entropy rate is represented mathematically as

                       \frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma  = 0

          =>          \frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)

    generally (s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]

substituting for (s_1 -s_2)

                      \frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} +  c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]

                      Where \frac{\sigma}{\r m} is the rate of entropy produced within the turbine

 substituting values

                \frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]

                    \frac{\sigma}{\r m}=  0.0861kJ/kg \cdot K

           

 

                   

   

5 0
4 years ago
"Using your favorite search engine and the resources of your library, develop a set of recommendations regarding the possible us
alina1380 [7]

Answer:

Cloud computing services are going to be very important to supporting Ashville's mobile app since it has different uses. These applications uses include; data storage, as well as helping to provide network to any business data related work. This will improve the city of Ashville's mobile app operations considerably.

The benefits of using cloud computing are; it allows a business managers to be more concerned with running the business themselves rather than maintaining the data center. The use of cloud computing allows a company's IT administrators to focus on managing the company's operations, thereby allowing performance to be enhanced of the business enterprise.

A commercial company is able to create new technologies more rapidly with the use of cloud computing. Furthermore, the enterprise will be able to automate its activities using cloud computing. Cloud computing is also very important as it is more affordable and thus promotes the company's growth in the market. Cloud computing is also enhancing the global presence of the Ashville mobile app. The drawbacks of using cloud computing include the following; due to using the cloud infrastructure, the company's performance may be unreliable. People believe that cloud computing is not reliable and that cloud computing may not be secure at last and can't always be right for all workloads.

Explanation:

5 0
3 years ago
The 5-kg collar has a velocity of 5 m>s to the right when it is at A. It then travels along the smooth guide. Determine its s
Gnoma [55]

Answer:

The speed at point B is 5.33 m/s

The normal force at point B is 694 N

Explanation:

The length of the spring when the collar is in point A is equal to:

lA=\sqrt{0.2^{2}+0.2^{2}  }=0.2\sqrt{2}m

The length in point B is:

lB=0.2+0.2=0.4 m

The equation of conservation of energy is:

(Tc+Ts+Vc+Vs)_{A}=(Tc+Ts+Vc+Vs)_{B} (eq. 1)

Where in point A: Tc = 1/2 mcVA^2, Ts=0, Vc=mcghA, Vs=1/2k(lA-lul)^2

in point B: Ts=0, Vc=0, Tc = 1/2 mcVB^2, Vs=1/2k(lB-lul)^2

Replacing in eq. 1:

\frac{1}{2}m_{c}v_{A}^{2}+0+m_{c}gh_{A}+      \frac{1}{2}k(l_{A}-l_{ul})  ^{2}=\frac{1}{2}m_{c}v_{B}^{2}+0+0+\frac{1}{2}k(l_{B}-l_{ul})  ^{2}

Replacing values and clearing vB:

vB = 5.33 m/s

The balance forces acting in point B is:

Fc-NB-Fs=0

\frac{m_{C}v_{B}^{2}   }{R}-N_{B}-k(l_{B}-l_{ul})=0

Replacing values and clearing NB:

NB = 694 N

6 0
3 years ago
Read 2 more answers
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