Sea A una matriz 3x3 con la propiedad de que la transformada lineal x → Ax mapea R³ sobre R³.

if when you put your shirt in your pants, your shirt is tucked, does that mean when your shirt is over your pants, your pants ar

Archy [21]

Answer:

confusing, but yes

Explanation:

8 0

3 years ago

Read 2 more answers

A cylinder contains 480 cm3 of loose dry sand which weighs 820 g. Under a static load of 200 kPa the volume is reduced 1%, and t

goblinko [34]

Answer:

a.

b.

c.

Explanation:

a. void ratio is provided by the formula: $e = \frac{V_{p} }{V_{s} }$

where , $V_{p}$ = volume of voids

$V_{s}$ = volume of solid grains

for loose sand, the void space = $\frac{480}{480}$

= 1

b. void ratio after static load = 0.1/(480)/ (480)

= 0.1

c. void ratio after vibration = [480- ( 0.1 * 480) ]/ 480

= 0.9

5 0

3 years ago

For a given set of input values, a NAND gate produces the opposite output as an OR gate with inverted inputs.A. True

Verdich [7]

Answer:

B.

Explanation:

For a given set of input values, A NAND gate produces exactly the same values as an OR gate with inverted inputs.

The truth table for a NAND gate with 2 inputs is as follows:

0 0 1

0 1 1

1 0 1

1 1 0

The truth table for an OR gate, is as follows:

0 0 0

0 1 1

1 0 1

1 1 1

If we add two extra columns for inverted inputs, the truth table will be this one:

0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

1 1 0 0 0

which is the same as for the NAND gate, not the opposite, so the statement is false.

This means that the right choice is B.

3 0

3 years ago

Pipe (2) is supported by a pin at bracket C and by tie rod (1). The structure supports a load P at pin B. Tie rod (1) has a diam

Galina-37 [17]

Answer:

P_max = 25204 N

Explanation:

Given:

- Rod 1 : Diameter D = 12 mm , stress_1 = 110 MPa

- Rod 2: OD = 48 mm , thickness t = 5 mm , stress_2 = 65 MPa

- x_1 = 3.5 mm ; x_2 = 2.1 m ; y_1 = 3.7 m

Find:

- Maximum Force P_max that this structure can support.

Solution:

- We will investigate the maximum load that each Rod can bear by computing the normal stress due to applied force and the geometry of the structure.

- The two components of force P normal to rods are:

Rod 1 : P*cos(Q)

Rod 2: - P*sin(Q)

where Q: angle subtended between x_1 and Rod 1 @ A. Hence,

Q = arctan ( y_1 / x_1)

Q = arctan (3.7 / 2.1 ) = 60.422 degrees.

- The normal stress in each Rod due to normal force P are:

Rod 1 : stress_1 = P*cos(Q) / A_1

Rod 2: stress_2 = - P*sin(Q) / A_2

- The cross sectional Area of both rods are A_1 and A_2:

A_1 = pi*D^2 / 4

A_2 = pi*(OD^2 - ID^2) / 4

- The maximum force for the given allowable stresses are:

Rod 1: P_max = stress_1 * A_1 / cos(Q)

P_max = (110*10^6)*pi*0.012^2 / 4*cos(60.422)

P_max = 25203.61848 N

Rod 2: P_max = stress_2 * A_2 / sin(Q)

P_max = (65*10^6)*pi*(0.048^2 - 0.038^2) / 4*sin(60.422)

P_max = 50483.4 N

- The maximum force that the structure can with-stand is governed by the member of the structure that fails first. In our case Rod 1 with P_max = 25204 N.

8 0

3 years ago

(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th

arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation ( $\dot Q$ ), measured in BTU per hour, is represented by this formula:

$\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4})$ (1)

Where:

$\epsilon$ - Emissivity, dimensionless.

$A$ - Surface area of the student, measured in square feet.

$\sigma$ - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

$T_{s}$ - Temperature of the student, measured in Rankine.

$T_{b}$ - Temperature of the bus, measured in Rankine.

If we know that $\epsilon = 0.90$ , $A = 16.188\,ft^{2}$ , $\sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}$ , $T_{s} = 554.07\,R$ and $T_{b} = 527.67\,R$ , then the heat transfer rate due to electromagnetic radiation is:

$\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]$