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sasho [114]
3 years ago
14

True or false tensile forces are smaller in arch bridges

Engineering
1 answer:
Ivan3 years ago
4 0

Answer:

True

Explanation:

The tensile forces are small in most arches and usually negligible.

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Using your knowledge of how an ATM is used, develop a set of use-cases that could serve as a basis for understanding the require
drek231 [11]

Answer:

Use cases are known to be a set of instruction  or processes between a User/Actor with the system to produce a desired input.

With the aid of a diagram, the set of use cases that are carried out in this ATM are given below:

Insert PIN

(1)Perform required transaction

(2)Withdrawal

(3)Deposit

(4)Transfer

(5)Change PIN

(6)Exit

Note: Kindly find an attached diagram of the Use case as part of the solution to process carried out at the ATM

Sources: The diagram of the Use case for ATM was researched and taken from Quizlet.

Explanation:

Solution

Use cases are normally a set of instruction  or processes between a User/Actor with the system to produce a desired input.

A use case diagram or image is a graphical representation of all the use case or processes that connects or interact with the system

The use case diagram is a part of Unified Modelling Language also called the UML.

The set of use cases that are carried out in this ATM use case diagram to know the requirements of the ATM is shown below:

  • Insert PIN
  • Perform required transaction
  • Withdrawal
  • Deposit
  • Transfer
  • Change PIN
  • Exit

Now both the customer/client and Bank are seen as Actors.

Actors are the ones or people that interface with the system.

6 0
3 years ago
Which option shows the most valuable metallic properties
Rina8888 [55]

Malleable and ductile

non metals like plastic also have other properties but can't be malleable and ductile so they r most valuable metallic properties

6 0
2 years ago
me that both a triaxial shear test and a direct shear test were performed on a sample of dry sand. When the triaxial test is per
stepan [7]

Answer:

shear strength = 2682.31 Ib/ft^2

Explanation:

major principal stress = 100 Ib / in2

minor principal stress = 20 Ib/in2

Normal stress = 3000 Ib/ft2

<u>Determine the shear strength when direct shear test is performed </u>

To resolve this we will apply the coulomb failure criteria relationship between major and minor principal stress a

for direct shear test

use Mohr Coulomb criteria relation between normal stress and shear stress

Shear strength when normal strength is 3000 Ib/ft  = 2682.31 Ib/ft^2

attached below is the detailed solution

8 0
3 years ago
Oliver is designing a new children's slide to increase speed at which a child can descend.His first design involved steel becaus
Ede4ka [16]

Answer:

Steel can rusts easily in the presence of moisture and oxygen, and tarnishes as the rust progresses.

Explanation:

Steel is an alloy of carbon and iron. it is a very useful alloy that is found in almost any engineered piece. The problem with steel is that it is very susceptible to rust, and the cost of maintaining it in order to prevent rust is very high. Steel rusts in the presence of moisture and oxygen (rust is an oxidation-reduction process). Using it as a water slide exposes it constantly to moisture, and to prevent it from rusting in this case will involve a lot of maintenance cost, which is why steel is not advisable to be used in this case.

4 0
3 years ago
An automobile has a mass of 1200 kg. What is its kinetic energy, in ki, relative to the road when traveling at a velocity of 50
netineya [11]

Answer:

a)Ek=115759.26J

b)23.15kW

Explanation:

kinetic energy(Ek) is understood as that energy that has a body with mass when it travels with a certain speed, it is calculated by the following equation

Ek=0.5mv^{2}

for the first part

m=1200Kg

V=50km/h=13.89m/s

solving for Ek

Ek=0.5(1200)(13.89)^2

Ek=115759.26J

For the second part of the problem we calculate the kinetic energy, using the same formula, then in order to find the energy change we find the difference between the two, finally divide over time (15s) to find the power.

m=1200kg

V=100km/h=27.78m/s

Ek=0.5(1200)(27.78)^2=462963J\\

taking into account all of the above the following equation is inferred

ΔE=\frac{Ek2-EK1}{T}=\frac{462963-115759.26}{15}  =23146.916W=23.15kW

3 0
3 years ago
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