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2 years ago

Which of the following statements is not necessarily true for a well-arranged floor plan?

1 answer:

4 0

A well arranged floor plan is one that optimises the given floor area. Floor plans are useful to help design furniture layout, wiring systems, and much more. Option B is the answer, since it does not meet the standard of a plan

<h3>What is a Floor Plan?</h3>

A floor plan is a scaled diagram of a room or building viewed from above. The floor plan may depict an entire building, one floor of a building, or a single room. It may also include measurements, furniture, appliances, or anything else necessary to the purpose of the plan.

<h3>Other properties of a floor plan are:</h3>

Maximize the property
Utilize space effectively
Accessibility
Flexibility
Functionality
Maximize the use of light
Attention to size
Fitting to your lifestyle

Learn more:

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The best way to identify common masonry problems is to call the engineer.<br> True or False

Daniel [21]

Answer:

True

Explanation:

5 0

3 years ago

A well is located in a 20.1-m thick confined aquifer with a conductivity of 14.9 m/day and a storativity of 0.0051. If the well

ahrayia [7]

Answer:

S = 5.7209 M

Explanation:

Given data:

B = 20.1 m

conductivity ( K ) = 14.9 m/day

Storativity ( s ) = 0.0051

1 gpm = 5.451 m^3/day

calculate the Transmissibility ( T ) = K * B

= 14.9 * 20.1 = 299.5 m^2/day

Note :

t = 1

U = ( r^2* S ) / (4*T*<em> t </em>)

= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4

Applying the thesis method

W(u) = -0.5772 - In(U)

= 7.9

next we calculate the pumping rate from well ( Q ) in m^3/day

= 500 * 5.451 m^3 /day

= 2725.5 m^3 /day

Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping

S = $\frac{Q}{4\pi T} * W (u)$

where : Q = 2725.5

T = 299.5

W(u) = 7.9

substitute the given values into equation above

S = 5.7209 M

4 0

3 years ago

A 1 m3 rigid tank initially contains air whose density is 1.18kg/m3. The tank is connected to a high pressure supply line throug

Elanso [62]

To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.

Density can be defined as

$\rho = \frac{m}{V}$

Where

m = Mass

V = Volume

For state one we know that

$\rho_1 = \frac{m_1}{V}$

$m_1 = \rho_1 V$

$m_1 = 1.18*1$

$m_1 = 1.18Kg$

For state two we have to

$\rho_2 = \frac{m_2}{V}$

$m_2 = \rho_2 V$

$m_1 = 7.2*1$

$m_1 = 7.2Kg$

Therefore the total change of mass would be

$\Delta m = m_2-m_1$

$\Delta m = 7.2-1.18$

$\Delta m = 6.02Kg$

Therefore the mass of air that has entered to the tank is 6.02Kg

5 0

3 years ago

When Ontario

Marat540 [252]

Answer:

Explanation:

This will be possible when setting them up in summer with a certain quantity of sag, they have already know that the cables won't be able to sag any further because of the heat. During winter, when the cables contract because of the cold weather, the sag will therefore be reduced, but much tension will not be put on the cables.

4 0

2 years ago

An AM radio transmitter operating at 3.9 MHz is modulated by frequencies up to 4 kHz. What are the maximum upper and lower side

Aliun [14]

Answer:

total width bandwidth = 8kHz

Explanation:

given data

transmitter operating = 3.9 MHz

frequencies up to = 4 kHz

solution

we get here upper side frequencies that is

upper side frequencies = 3.9 × $10^{6}$ + 4 × 10³

upper side frequencies = 3.904 MHz

and

now we get lower side frequencies that is

lower side frequencies = 3.9 × $10^{6}$ - 4 × 10³

lower side frequencies = 3.896 MHz

and now we get total width bandwidth

total width bandwidth = upper side frequencies - lower side frequencies

total width bandwidth = 8kHz

6 0

3 years ago