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GrogVix [38]
3 years ago
10

For this problem, calculate the following by hand and show the procedure for how you obtained the results. Subsequently, solve p

arts (a-d) using MATLAB and include your MATLAB code in your submission. a) Determine the inner product of vectors a and b. b) Determine the outer (cross) product of vectors a and b. c) Calculate norm (magnitude) of vector c. d) Calculate the determinant of 3x3 matrix A, in which the first column is vector a, second column is vector b, and the third column is vector c. e) Determine if vectors a, b and c are linearly independent.
Engineering
1 answer:
Gemiola [76]3 years ago
4 0

Answer:wat

Explanation:

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The input and output signals of a system is related by the following equation: fraction numerator d squared y over denominator d
Colt1911 [192]

Answer:

Explanation:

The given equation is :

\frac{d^{2}y }{dx^{2} } + sin(3y) \frac{dy}{dt} + y = t\frac{df}{dt} + f

5 0
3 years ago
A machine used to shred cardboard boxes for composting has a first cost of $10,000, an AOC of $7000 per year, a 3-year life, and
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Annual Payment where F is accumulated sum of amount, n is number of years and i is annual rate of interest. The standard notation equation is in the image since i can’t type it-
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3 years ago
A cylindrical insulation for a steam pipe has an inside radius rt = 6 cm, outside radius r0 = 8 cm, and a thermal conductivity k
goldfiish [28.3K]

Answer:

heat loss per 1-m length of this insulation is 4368.145 W

Explanation:

given data

inside radius r1 = 6 cm

outside radius r2 = 8 cm

thermal conductivity k = 0.5 W/m°C

inside temperature t1 = 430°C

outside temperature t2 = 30°C

to find out

Determine the heat loss per 1-m length of this insulation

solution

we know thermal resistance formula for cylinder that is express as

Rth = \frac{ln\frac{r2}{r1}}{2 \pi *k * L}   .................1

here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity

so

heat loss is change in temperature divide thermal resistance

Q = \frac{t1- t2}{\frac{ln\frac{r2}{r1}}{2 \pi *k * L}}

Q = \frac{(430-30)*(2 \pi * 0.5 * 1}{ln\frac{8}{6} }

Q = 4368.145 W

so heat loss per 1-m length of this insulation is 4368.145 W

4 0
3 years ago
A 1000 W iron utilizes a resistance wire which is 20 inches long and has a diameter of 0.08 inches. Determine the rate of heat g
SSSSS [86.1K]

Answer:

The rate of heat generation in the wire per unit volume is 5.79×10^7 Btu/hrft^3

Heat flux is 9.67×10^7 Btu/hrft^2

Explanation:

Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr

Area (A) = πD^2/4

Diameter (D) = 0.08 inches = 0.08 in × 3.2808 ft/39.37 in = 0.0067 ft

A = 3.142×0.0067^2/4 = 3.53×10^-5 ft^2

Volume (V) = A × Length

L = 20 inches = 20 in × 3.2808 ft/39.37 in = 1.67 ft

V = 3.53×10^-5 × 1.67 = 5.8951×10^-5 ft^3

Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr ÷ 5.8951×10^-5 ft^3 = 5.79×10^7 Btu/hrft^3

Heat flux = 3412.15 Btu/hr ÷ 3.53×10^-5 ft^2 = 9.67×10^7 Btu/hrft^2

3 0
3 years ago
Resistance to impact is an example of a(n)
Anika [276]

Answer:

Mechanical property

Explanation:

MECHANICAL PROPERTIES can be defined as the ability of a metal or material to remain undamaged after different type of forces has been applied or used on them because forces or loads are often applied to metal, material or physical properties which is why MECHANICAL PROPERTIES enables us to know the strength , toughness as well as the hardness of metal and the way this metal perform or react when different forces are applied on them.

Lastly any metal, material or physical properties that has the strength , hardness and resistance to withstand or remain unaffected despite the loads or forces use on them is an example of MECHANICAL PROPERTIES.

Therefore Resistance to impact is an example of a(n) MECHANICAL PROPERTIES.

8 0
3 years ago
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