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3 years ago

By conducting a survey, Sarah and Jesse determine that an “open space” is home to over fifty different species of plant life.

Engineering

2 answers:

stellarik [79]3 years ago

7 0

Answer: True

Many animals and plants live in the areas that aren´t occupied by human civiliaztions.

Taya2010 [7]3 years ago

3 0

I feel as though there isn’t enough information to state if it’s true or false because we don’t know where the open space is and how many things may be in the open space

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A certain working substance receives 100 Btu reversibly as heat at a temperature of 1000℉ from an energy source at 3600°R. Refer

Valentin [98]

Answer:

Explanation:

t1 = 1000 F = 1460 R

t0 = 80 F = 540 R

T2 = 3600 R

The working substance has an available energy in reference to the 80F source of:

B1 = Q1 * (1 - T0 / T1)

B1 = 100 * (1 - 540 / 1460) = 63 BTU

The available energy of the heat from the heat wource at 3600 R is

B2 = Q1 * (1 - T0 / T2)

B2 = 100 * (1 - 540 / 3600) = 85 BTU

The reduction of available energy between the source and the 1460 R temperature is:

B3 = B2 - B1 = 85 - 63 = 22 BTU

6 0

3 years ago

Pls pls pls pls plsss someone help pls

snow_tiger [21]

Answer:

I would like to help can you take a picture without the bot telling you something. Maybe then I can figure it out.

6 0

2 years ago

A stainless steel ball (=8055 kg/m3, Cp= 480 J/kgK) of diameter D =15 cm is removed from theoven at a uniform temperature of 3

aleksandrvk [35]

Answer:

i) 25.04 W/m^2 .k

ii) 23.82 minutes = 1429.2 secs

Explanation:

Given data:

Diameter of steel ball = 15 cm

uniform temperature = 350°C

p = 8055 kg/m^3

Cp = 480 J/kg.k

surface temp of ball drops to 250°C

average surface temperature = ( 350 + 250 ) / 2 = 300°C

i) Determine the average convection heat transfer coefficient during the cooling process

Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table " properties of air " contained in your textbook

average convection heat transfer coefficient = 25.04 W/m^2 .k

ii) Determine how long this process has taken

Time taken by the process = 23.82 minutes = 1429.2 seconds

Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs

attached below is the detailed solution of the given question

3 0

3 years ago

A strain gauge with a 4 mm gauge length gives a displacement reading of 1.5 um. Calculate the stress at the location of the stra

Art [367]

Answer:

1) 75Mpa

2) 1.125 MPa

Explanation:

given data:

gauge length = 4 mm

displacement $= 1.5\mu m = 1.5\times 10^{-3} m$

a) structural steel

Young modulus for steel is $200 GPa = 200\times 10^3 MPa$

we know that

$E =\frac{stress}{strain}$

$Stress = 200\times 10^3 \times \frac{1.5\times 10^{-3}}{4}$

= 75Mpa

b) PMMA

Young's modulus $= 3GPa = 3\times10^3 MPa$

$stress = 3\times \frac{1.5\times10^{-3}}{4}$

stress = 1.125 MPa

3 0

3 years ago

BCC lithium has a lattice parameter of 3.5089 3 10–8 cm and contains one vacancy per 200 unit cells. Calculate (a) the number of

Tanya [424]

(a) The number of vacancies per cubic centimeter is 1.157 X 10²⁰

(b) ρ = n X (AM) / v X Nₐ

Explanation:

Given-

Lattice parameter of Li = 3.5089 X 10⁻⁸ cm

1 vacancy per 200 unit cells

Vacancy per cell = 1/200

(a)

Number of vacancies per cubic cm = ?

Vacancies/cm³ = vacancy per cell / (lattice parameter)³

Vacancies/cm³ = 1 / 200 X (3.5089 X 10⁻⁸cm)³

Vacancies/cm³ = 1.157 X 10²⁰

Therefore, the number of vacancies per cubic centimeter is 1.157 X 10²⁰

(b)

Density is represented by ρ

ρ = n X (AM) / v X Nₐ

where,

Nₐ = Avogadro number

AM = atomic mass

n = number of atoms

v = volume of unit cell

4 0

3 years ago