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4 years ago

Engineering Careers Scavenger Hunt

Engineering

2 answers:

emmasim [6.3K]4 years ago

6 0

Answer:

Explanation:

it's the only engineering career

tia_tia [17]4 years ago

5 0

C is the only engineering career :)

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Find the total present worth of a series of cash flows with an annual interest rate of 2% per year. Round your answer to the nea

prisoha [69]

The total present worth is $19,783.01

The present worth of a series of cash flow is the value of the cash flows in year 0 (today)

Cash flow in year 0 = 5330

Cash flow in year 1 = 0

Cash flow in year 2 = 0

Cash flow in year 3 = 13075 / (1.02)^3 = 12,320.86

Cash flow in year 4 = 2308 / (1.02)^4 = 2,132.24

Present worth = $19,783.01

A similar question was solved here: brainly.com/question/9641711?referrer=searchResults

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3 years ago

The steel 4140 steel contains 0.4% C, however, it shows higher yield strength and ultimate strength than that of the 1045 (0.45%

Aleonysh [2.5K]

Answer:

4140 steel contains 0.4% C having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

Explanation:

we have given 4140 steel contains 0.4% C

we know here that 4140 steel is low steel alloy , and it have low amount of chromium , manganese etc alloying element

and these elements which are present in 4140 steel they increase yield strength and ultimate strength of steel

while in 1045 steel contains 0.45 % c is plain carbon steel

and it do not contain any alloying element

so that 4140 steel contains 0.4% C having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

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3 years ago

Turning a screw with a screwdriver pushes the screw into the wood. The force is __________ over many turns

IrinaK [193]

Answer:

repeated?

Explanation:

not really sure what type of answer choices you have

5 0

3 years ago

Whats viruses c liver?

PtichkaEL [24]

Answer:

Hepatitis C is a viral infection that causes liver inflammation, sometimes leading to serious liver damage. The hepatitis C virus (HCV) spreads through contaminated blood.

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2 years ago

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A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °

SOVA2 [1]

Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30 (Btu/hr)/(ft ⋅ °F)= $\frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3} (Btu/s)/(ft.F)$

Thermal Conductivity is SI units:

$k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K$

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

$A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2$

Heat Q is:

$Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J$

In Btu:

$A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2$

Heat Q is:

$Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu$

PArt B:

At midpoint Length=L/2=0.1 m

$Q=\frac{k*A*(T_1-T_2)}{L}$

On rearranging:

$T_2=T_1-\frac{Q*L}{KA}$

$T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C$

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3 years ago