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love history [14]
3 years ago
15

Please help me fast, I don’t have time

Engineering
1 answer:
Anna71 [15]3 years ago
5 0

Answer: precision

Explanation: Because accuracy is where you keep on getting it right but precision is where you get closer and closer

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Calculate the theoretical density of FCC iron (eg. austenitic stainless steel). The lattice parameter for FCC iron is 0.357 nm a
Ann [662]

Answer: 12.4 feet

Explanation:

If there is a smooth transition and there is no change in slopes, energy considerations can be used

The cube has a kinetic energy of

ke = mv^2/2 = 10 lbm * 20^2ft^2/s^2 / 2 = 2000 lbm-ft^2 / s^2

At the highest point when there is a gain in potential energy

pe = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2

If there is no loss in energies,

pe = ke

322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2

h = 2000 /322 = 6.211 (ft)

= h / sin(30) = 12.4 ft

8 0
3 years ago
Given x = 67 and y = 18, use two's complement to calculate:(1)x+y (2) x-y (3)-x+y (4)-x-y. Do the calculation, show the results
joja [24]

Answer:

See attachment

Explanation:

8 0
3 years ago
Which of the following are three criteria for creating measurable product attributes?
Vinil7 [7]

Answer:

it A

Explanation:

8 0
3 years ago
Read 2 more answers
The less surface area of gasoline exposed to the air, the faster a given amount will burn. True of False
romanna [79]

Answer:

true

Explanation:

it depends on how fast the car goes because the gas burns faster or slower depending on the speed you go or drive. if your drive 50 MPH, the gas will burn slowly. if your drive 45 MPH, the gas will burn faster if you go slow too much. if you press the glass button and hold it for 5 minutes, the more you hold the gas for too long, the more the gas will burn inside the gas engine. go look at your car or your parents car and see how it goes and that will help.

4 0
3 years ago
A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 4.5
IgorC [24]

Answer:

The maximum length of the specimen before the deformation was 358 mm or 0.358 m.

Explanation:

The specific deformation ε for the material is:

\epsilon = \deltaL /L (1)

Where δL and L represent the elongation and initial length respectively. From the HOOK's law we also now that for a linear deformation, the deformation and the normal stress applied relation can be written as:

\sigma = E/ \epsilon (2)

Where E represents the elasticity modulus. By combining equations (1) and (2) in the following form:

L= \delta L/ \epsilon

\epsilon =\sigma /E  

So by calculating ε then will be possible to find L. The normal stress σ is computing with the applied force F and the cross-sectional area A:

\sigma=F/A

\sigma=\frac{F} {\pi*d^2/4}

\sigma=\frac{2170 N}{\pi*4.5 mm^2/4}

\sigma= 136000000 Pa= 136 Mpa  

Then de specific defotmation:

\epsilon =136 MPa / 108 GPa = 1.26*10^{-3}

Finally the maximum specimen lenght for a elongation 0f 0.45 mm is:

L= 0.45 mm/ 1.26*10^{-3} = 358 mm = 0.358 m

4 0
3 years ago
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