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Kisachek [45]
3 years ago
15

Given int variables k and total that have already been declared, use a do...while loop to compute the sum of the squares of the

first 50 counting numbers, and store this value in total. Thus your code should put 1*1 + 2*2 + 3*3 +... + 49*49 + 50*50 into total. Use no variables other than k and total.
Engineering
1 answer:
Otrada [13]3 years ago
5 0

Answer:

k-1000

Explanation:

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Who play 1v1 lol unblocked games 76
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Read 2 more answers
Consider the diffusion of water vapor through a polypropylene (PP) sheet 1 mm thick. The pressures of H2O at the two faces are 3
Neko [114]

Answer:

\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}

Explanation:

Diffusion flux of a gas, J is given by

J=P_m\frac {\triangle P}{\triangle x} where P_m is permeability coefficient, \triangle P is pressure difference and x is thickness of membrane.

The pressure difference will be 10,000 Pa- 3000 Pa= 7000 Pa

At 298 K, the permeability coefficient of water vapour through polypropylene sheet is 38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)

Since the thickness of sheet is given as 1mm= 0.1 cm then

J=38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)\times \frac {7000 pa}{0.1cm}=0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}

Therefore, the diffusion flux is \boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}

7 0
3 years ago
Four common causes of product failure are poor design, poor construction, poorly communicated operating instructions, and ______
Inga [223]

Answer:

B. operator error or misuse

Explanation:

A product is a failure if it is not able to achieve the anticipated life cycle as expected by the organization.

In such a case, there is a withdrawal of the product from the market as a result of its ultimate failure of a product to achieve profitability.

Four common causes of product failure are poor design, poor construction, poorly communicated operating instructions, and <u>operator error or misuse</u>

3 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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