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4 years ago

; How do engineers make a difference in the world and with proof ?

2 answers:

6 0

Well Engineers make a difference in the world because well for an example when u buy a car or something like that someone has made it because if u had no car were would u go if u had to go somewhere like New York u can’t walk from there and back u would be exhausted and die so with things like car, trucks, and bus’s u wouldn’t have something to get u to point A to B and maybe C that if u have a point c well hoped it helped

kumpel [21]4 years ago

3 0

Engineers are the reason healthcare has improved so dramatically throughout the years. The advancements in medical technology are down to the work conducted by engineers as well as the creation of devices that help to save lives and improve the quality of life for others. New developments are being created by engineers constantly, for example, the Crossrail project in London is creating a new transport system throughout the South East to create shorter journey times and easier connections. The existing transport system wouldn’t be where it is today without engineers.

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Horizontal shear forces and, consequently, horizontal shear stresses are caused in a flexural member at those locations where th

jek_recluse [69]

Answer:

False

Explanation:

When the horizontal shear forces act on the surface there is transverse shear stress at a particular point which is equal in magnitude. Pure bending is less common than a non uniform bending because the beam is not in equilibrium.

5 0

3 years ago

A rectangular bar of length L has a slot in the central half of its length. The bar has width b, thickness t, and elastic modulu

Naya [18.7K]

Answer:

The correct answer to the following question will be "1.23 mm".

Explanation:

The given values are:

Average normal stress,

$\sigma=200 \ MPa$

Elastic module,

$E = 77 \ GPa$

Length,

$L = 570 \ mm$

To find the deformation, firstly we have to find the equation:

⇒ $\delta=\Sigma\frac{N_{i}L_{i}}{E \ A_{i}}$

⇒ $=\frac{P(\frac{L}{H})}{E(bt)} +\frac{P(\frac{L}{2})}{E (bt)(\frac{2}{3})}+\frac{P(\frac{L}{H})}{Ebt}$

On taking " $\frac{PL}{Ebt}$ " as common, we get

⇒ $=\frac{\frac{PL}{Ebt}}{[\frac{1}{4}+\frac{3}{4}+\frac{1}{4}]}$

⇒ $=\frac{5PL}{HEbt}$

Now,

The stress at the middle will be:

⇒ $\sigma=\frac{P}{A}$

⇒ $=\frac{P}{(\frac{2}{3})bt}$

⇒ $=\frac{3P}{2bt}$

⇒ $\frac{P}{bt} =\frac{2 \sigma}{3}$

Hence,

⇒ $\delta=\frac{5 \sigma \ L}{6E}$

On putting the estimated values, we get

⇒ $=\frac{5\times 200\times 570}{6\times 77\times 10^3}$

⇒ $=\frac{570000}{462000}$

⇒ $=1.23 \ mm$

8 0

3 years ago

A customer got debited twice, how would you investigate and resolve this conflict on the customer side and on the company side?

eimsori [14]

Answer:

First, the question seems to assume that the debit is erroneous. Because, if it's intentional, then there is little need for an investigation.

So if there was an unintentional duplication of transaction, this would totally depend on what kind of debit it is. So to answer the question, we'll explore as many scenarios as possible.

Debit on a Bank Account (Business to Bank)

If it is a debit on the bank account, then the statement of account from the company side and the account activity(ies) leading up to the debits will reveal why there is a double debit. It is possible the transaction was entered twice. This can happen when there is a bad connection that creates a failed transaction message on the customer's end but still carries through with the payment instruction. In this case, the customer will think the debit or payment request didn't go through and proceed to repeat the same transaction.

Banks do execute debits on accounts for charges such as Bank Account Maintenance Charges, Credit card charges etc. Errors could also arise internally that led to the duplication of a debit instruction either due to a bug in the system.

So in this case, we'll need to access the customer's account internally whilst externally, the client will need to provide the message containing the alerts, date and time of debits etc. to help the banks further the investigation.

Business to Business

This could arise out of a "failed POS" transaction. Similar to the condition stated above, when there is a glitch with the network and the POS fails, it should print out a receipt that reads failed or declined.

This receipt will contain references to the transaction. This will aid the investigation of the debits to the account.

Explanation:

8 0

3 years ago

A mass of air occupying a volume of 0.15m^3 at 3.5 bar and 150 °C is allowed [13] to expand isentropically to 1.05 bar. Its enth

e-lub [12.9K]

Answer:

Total work: -5.25 kJ

Total Heat: 52 kJ

Explanation:

V0 = 0.15

P0 = 350 kPa

t0 = 150 C = 423 K

P1 = 105 kPa (isentropical transformation)

Δh1-2 = 52 kJ (at constant pressure)

Ideal gas equation:

P * V = m * R * T

m = (R * T) / (P * V)

R is 0.287 kJ/kg for air

m = (0.287 * 423) / (350 * 0.15) = 2.25 kg

The specifiv volume is

v0 = V0/m = 0.15 / 2.25 = 0.067 m^3/kg

Now we calculate the parameters at point 1

T1/T0 = (P1/P0)^((k-1)/k)

k for air is 1.4

T1 = T0 * (P1/P0)^((k-1)/k)

T1 = 423 * (105/350)^((1.4-1)/1.4) = 300 K

The ideal gas equation:

P0 * v0 / T0 = P1 * v1 / T1

v1 = P0 * v0 * T1 / (T0 * P1)

v1 = 350 * 0.067 * 300 / (423 * 105) = 0.16 m^3/kg

V1 = m * v1 = 2.25 * 0.16 = 0.36 m^3

The work of this transformation is:

L1 = P1*V1 - P0*V0

L1 = 105*0.36 - 350*0.15 = -14.7 kJ/kg

Q1 = 0 because it is an isentropic process.

Then the second transformation. It is at constant pressure.

P2 = P1 = 105 kPa

The enthalpy is raised in 52 kJ

Cv * T1 + P1*v1 = Cv * T2 + P2*v2 + Δh

And the idal gas equation is:

P1 * v1 / T1 = P2 * v2 / T2

T2 = T1 * P2 * v2 / (P1 * v1)

Replacing:

Cv * T1 + P1*v1 + Δh = Cv * T1 * P2 * v2 / (P1 * v1) + P2*v2

Cv * T1 + P1*v1 + Δh = v2 * (Cv * T1 * P2 / (P1 * v1) + P2)

v2 = (Cv * T1 + P1*v1 + Δh) / (Cv * T1 * P2 / (P1 * v1) + P2)

The Cv of air is 0.7 kJ/kg

v2 = (0.7 * 300 + 105*0.16 + 52) / (0.7 * 300 * 105 / (105 * 0.16) + 105) = 0.2 m^3/kg

V2 = 2.25 * 0.2 = 0.45 m^3

T2 = 300 * 105 * 0.2 / (105 * 0.16) = 375 K

The heat exchanged is Q = Δh = 52 kJ

The work is:

L2 = P2*V2 - P1*V1

L2 = 105 * 0.45 - 105 * 0.36 = 9.45 kJ

The total work is

L = L1 + L2

L = -14.7 + 9.45 = -5.25 kJ

8 0

3 years ago

A steel loop ABCD of length 5ft and of 3/8" diameter is placed as shown around a 1" diameter aluminum rod AC. Cables BE and DF e

lorasvet [3.4K]

Answer:

Explanation:

ye men thats the answer#######%

4 0

3 years ago

; How do engineers make a difference in the world and with proof ?​

; How do engineers make a difference in the world and with proof ?