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2 years ago

A(n) _____ is done to test each individual component (often a program) to ensure that it is as defect-free as possible.

Engineering

1 answer:

lyudmila [28]2 years ago

6 0

Answer:

Unit test

Explanation:

Defintion: A test of each individual component (often a program) to ensure that it is as defect-free as possible.

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A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a

nika2105 [10]

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, $g_{c}$ = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

$c_{p}$ = 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\$

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence $v_{a} ^{2} = 0$

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} + \frac{v_{1} ^{2} }{2c_{p} }$

$T_{1}$ = 20+460 = 480°R

$T_{a} =480+ \frac{(800)(800}{2(0.240)(25037}$ = 533.25°R

Pressure at the inlet of compressor at isentropic condition

$P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}$

$P_{a}$ = $(10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}$ = 14.45 psia

$P_{2}= 8P_{a} = 8(14.45) = 115.6 psia$

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3 years ago

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Answer:

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Explanation:

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3 years ago

Consider two different versions of algorithm for finding gcd of two numbers (as given below), Estimate how many times faster it

juin [17]

Answer:

Explanation:

Step 1:

a) The formula for compute greatest advisor is

gcd(m,n) = gcd (n,m mod n)

the gcd(31415,14142) by applying Euclid's algorithm is

gcd(31,415,14,142) =gcd(14,142,3,131)

=gcd=(3,131, 1,618)

=gcd(1,618, 1,513)

=gcd(1,513, 105)

=gcd(105, 43)

=gcd(43, 19)

=gcd(19, 5)

=gcd(5, 4)

=gcd(4, 1)

=gcd(1, 0)

STEP 2

b) The number of comparison of given input with the algorithm based on checking consecutive integers and Euclid's algorithm is

The number of division using Euclid's algorithm =10 from part (a)

The consecutive integer checking algorithm:

The number of iterations =14,142 and 1 or 2 division of iteration.

14,142 ∠= number of division∠ = 2*14,142

Euclid's algorithm is faster by at least 14,142/10 =1400 times

At most 2*14,142/10 =2800 times.

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3 years ago