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3 years ago

11

You want to improve your grades so that you make all A's next 6 weeks. List examples of quantitative and qualitative data you sh

ould collect. Suggest at least TWO ways to apply each CT strategy to the problem GIVE EXAMPLES and DETAILS!!!!!!!! You DO NOT simply define Plz help I will give brainliest

1 answer:

Naddika [18.5K]3 years ago

6 0

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

A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surf

galina1969 [7]

Answer:

Hello some parts of your question is missing below is the missing part

Convection coefficient = 11 w/m^2. °c

answer : 44.83 watts

Explanation:

Given data :

surface emissivity ( ε )= 0.95

head ( sphere) diameter( D ) = 0.25 m

Temperature of sphere( T ) = 35° C

Temperature of surrounding ( T∞ ) = 25°C

Temperature of surrounding surface ( Ts ) = 15°C

б = ( 5.67 * 10^-8 )

Determine the total rate of heat loss

First we calculate the surface area of the sphere

As = $\pi D^{2}$

= $\pi * 0.25^2$ = 0.2 m^2

next we calculate heat loss due to radiation

Qrad = ε * б * As( $T^{4} - T^{4} _{s}$ ) ---- ( 1 )

where ;

ε = 0.95

б = ( 5.67 * 10^-8 )

As = 0.2 m^2

T = 35 + 273 = 308 k

Ts = 15 + 273 = 288 k

input values into equation 1

Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )

= 22.83 watts

Qrad ( heat loss due to radiation ) = 22.83 watts

calculate the heat loss due to convection

Qconv = h* As ( ΔT )

= 11*0.2 ( 35 -25 ) = 22 watts

Hence total rate of heat loss

= 22 + 22.83

= 44.83 watts

5 0

3 years ago

What major advancement in machine tools occurred in the 1970s and what benefits did it provide? describe in your own words.

mixer [17]

Answer:

I'm just a seventh grader

4 0

3 years ago

Read 2 more answers

A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an

Anarel [89]

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here thermodynamic equation that

energy equation that is

$h1 + \frac{v1}{2} + q = h2 + \frac{v2}{2} + w$

put here value with

turbine is insulated so q = 0

so here

$3015.4 *1000 + \frac{10^2}{2} = 2431.7 * 1000 + \frac{90^2}{2} + w$

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

7 0

3 years ago

What are some advantages of making electronic components like transistors increasingly smaller?

Stells [14]

Answer:

Dr. Engelbart, who would later help develop the computer mouse and other personal computing technologies, theorized that as electronic circuits were made smaller, their components would get faster, require less power and become cheaper to produce — all at an accelerating pace

4 0

2 years ago