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3 years ago

6

Shania has started a new job as an app developer. Her first task was to make an old designed for Android available on other plat

forms. which of the following would make her job easiest and fastest?

1 answer:

LuckyWell [14K]3 years ago

7 0

The answer is here......

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Find the error in the following pseudo code

IRISSAK [1]

Pseudocodes are used as a prototype of an actual program.

The error in the pseudocode is that, the while loop in the pseudocode will run endlessly.

From the pseudocode, the first line is:

<em>Declare Boolean finished = false</em>

The while loop is created to keep running as long as <em>finished = false.</em>

So, for the while loop to end, the finished variable must be updated to true.

This action is not implemented in the pseudocode.

Hence, the error in the pseudocode is that, the while loop is an endless loop

Read more about pseudocodes at:

brainly.com/question/17442954

8 0

2 years ago

The ???? − i relationship for an electromagnetic system is given by ???? = 1.2i1/2 g where g is the air-gap length. For current

Artemon [7]

Answer:

a) The mechanical force is -226.2 N

b) Using the coenergy the mechanical force is -226.2 N

Explanation:

a) Energy of the system:

$\lambda =\frac{1.2*i^{1/2} }{g} \\i=(\frac{\lambda g}{1.2} )^{2}$

$\frac{\delta w_{f} }{\delta g} =\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

If i = 2A and g = 10 cm

$\lambda =\frac{1.2*i^{1/2} }{g} =\frac{1.2*2^{1/2} }{10x10^{-2} } =16.97$

$f_{m}=-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }=-\frac{16.97^{3}*2*0.1 }{3*1.2^{2} } =-226.2N$

b) Using the coenergy of the system:

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{1.2*2*i^{3/2} }{3*g^{2} }=-\frac{1.2*2*2^{3/2} }{3*0.1^{2} } =-226.2N$

8 0

3 years ago

The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas

Lilit [14]

Answer:

A) 209.12 GPa

B) 105.41 GPa

Explanation:

We are given;

Modulus of elasticity of the metal; E_m = 67 GPa

Modulus of elasticity of the oxide; E_f = 390 GPa

Composition of oxide particles; V_f = 44% = 0.44

A) Formula for upper bound modulus of elasticity is given as;

E = E_m(1 - V_f) + (E_f × V_f)

Plugging in the relevant values gives;

E = (67(1 - 0.44)) + (390 × 0.44)

E = 209.12 GPa

B) Formula for upper bound modulus of elasticity is given as;

E = 1/[(V_f/E_f) + (1 - V_f)/E_m]

Plugging in the relevant values;

E = 1/((0.44/390) + ((1 - 0.44)/67))

E = 105.41 GPa

4 0

3 years ago

The 240-ft structure is used to provide various support services to launch vehicles prior to liftoff. In a test, a 12-ton weight

alina1380 [7]

Answer:

hello your question lacks the required question attached below is the missing diagram

Forces in GJ = -4.4444 i.e. 4.4444 tons

Forces in IG = 15.382 tons ( T )

Explanation:

Forces in GJ = -4.4444 i.e. 4.4444 tons

Forces in IG = 15.382 tons ( T )

attached below is the detailed solution

3 0

3 years ago

Car insurance incentives and discounts are available depending on _____. A. school attendance and driver skill B. vehicle type a

borishaifa [10]

Answer:D. Location, vehicle type, and driving habits

7 0

4 years ago

Read 2 more answers