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3 years ago

U 4. Find 2 bridges in the US and answer the following:

Engineering

1 answer:

Zarrin [17]3 years ago

6 0

Answer:

Im guessing this is for CEA for PLTW, if so look up the exact assignment number and look at online examples of the exact same assignment.

Explanation:

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(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:

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

$\dot Q = 417.492\,\frac{BTU}{h}$

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

$Q = \dot Q \cdot \Delta t$ (2)

Where $\Delta t$ is the heat transfer time, measured in hours.

If we know that $\dot Q = 417.492\,\frac{BTU}{h}$ and $\Delta t = \frac{1}{3}\,h$ , then the net energy transfer is:

$Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)$

$Q = 139.164\,BTU$

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

7 0

3 years ago

Write a program that removes all spaces from the given input. You may assume that the input string will not exceed 50 characters

GrogVix [38]

Answer:

Program that removes all spaces from the given input

Explanation:

// An efficient Java program to remove all spaces

// from a string

class GFG

{

// Function to remove all spaces

// from a given string

static int removeSpaces(char []str)

{

// To keep track of non-space character count

int count = 0;

// Traverse the given string.

// If current character

// is not space, then place

// it at index 'count++'

for (int i = 0; i<str.length; i++)

if (str[i] != ' ')

str[count++] = str[i]; // here count is

// incremented

return count;

}

// Driver code

public static void main(String[] args)

{

char str[] = "g eeks for ge eeks ".toCharArray();

int i = removeSpaces(str);

System.out.println(String.valueOf(str).subSequence(0, i));

}

5 0

3 years ago

Read 2 more answers

A square silicon chip (k = 152 W/m·K) is of width 7 mm on a side and of thickness 3 mm. The chip is mounted in a substrate such

Harrizon [31]

Answer:

The steady-state temperature difference is 2.42 K

Explanation:

Rate of heat transfer = kA∆T/t

Rate of heat transfer = 6 W

k is the heat transfer coefficient = 152 W/m.K

A is the area of the square silicon = width^2 = (7/1000)^2 = 4.9×10^-5 m^2

t is the thickness of the silicon = 3 mm = 3/1000 = 0.003 m

6 = 152×4.9×10^-5×∆T/0.003

∆T = 6×0.003/152×4.9×10^-5 = 2.42 K

7 0

3 years ago

Acertain foundation will experience a bearing capacity failurewhen it is subjected to a downward load of 2200 kN. Using ASD with

ehidna [41]

Answer:

Um...

Explanation:

This is what I like to see teachers giving out.

7 0

3 years ago

Two resistors, A and B, individually connect to a 9V battery. A student notices that resistor A is warmer than resistor B. Which

dybincka [34]

Answer:

Resistor B

Explanation:

Since resistance is the opposition to the flow of current in a circuit,

first let assume the two resistors are connected in parallel to the voltage, recall that when connection is in parallel, the different amount of current pass through the resistors depending on the value with the small resistor having a lower resistance effect hence higher current will pass through

The energy dissipated in each resistor can be calculated as

$E=\frac{1}{2}IR^{2}t$ .

from the formula we can conclude that the energy value will be higher for the resistor with small resistance value. hence more heating effect which will cause it to be warm.

Also when connected individually the current flow from the voltage source will pass through the resistor which when we calculate the energy dissipated, the resistor with smaller value will be higher because it will draw more current which will in turn lead to a heating effect and cause the resistor to be warm. Hence we can conclude that the resistance B has greatest resistance value.

4 0

3 years ago