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2 years ago

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What is the average distance (in terms of R) between the mobile on the fringe of the serving cell and the second and third tier

of co-channel cells, and how many cells are in the second and third tier of co-channel cells for the cases of N= 1, N= 3, N= 4, N= 7, and N= 12 cluster sizes?

1 answer:

kramer2 years ago

4 0

Answer:

do you need all work shown for this?

You might be interested in

A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 4.5

IgorC [24]

Answer:

The maximum length of the specimen before the deformation was 358 mm or 0.358 m.

Explanation:

The specific deformation ε for the material is:

$\epsilon = \deltaL /L$ (1)

Where δL and L represent the elongation and initial length respectively. From the HOOK's law we also now that for a linear deformation, the deformation and the normal stress applied relation can be written as:

$\sigma = E/ \epsilon$ (2)

Where E represents the elasticity modulus. By combining equations (1) and (2) in the following form:

$L= \delta L/ \epsilon$

$\epsilon =\sigma /E$

So by calculating ε then will be possible to find L. The normal stress σ is computing with the applied force F and the cross-sectional area A:

$\sigma=F/A$

$\sigma=\frac{F} {\pi*d^2/4}$

$\sigma=\frac{2170 N}{\pi*4.5 mm^2/4}$

$\sigma= 136000000 Pa= 136 Mpa$

Then de specific defotmation:

$\epsilon =136 MPa / 108 GPa = 1.26*10^{-3}$

Finally the maximum specimen lenght for a elongation 0f 0.45 mm is:

$L= 0.45 mm/ 1.26*10^{-3} = 358 mm = 0.358 m$

4 0

2 years ago

What makes a tool a kind of technology?

ivolga24 [154]

Answer:

Option (C)

Explanation:

A tool is usually defined as an equipment or a device that is hold by the hands and is used to carry out a particular function. For example, a hammer is a tool that is used to hit on a hard surface.

A tool can be considered as a technology under a certain condition, such as the applications of this objects are helpful in solving a practical problem, with ease and comfortability.

Thus, the correct answer is option (C).

6 0

2 years ago

A 18-ft thick clay layer in the field (drained on one side) is normally consolidated. When the pressure is increase from 0.75 to

mihalych1998 [28]

Answer:

a) 294.34 days

b) Δh = 25.361 cm

Explanation:

Given data:

Thickness of clay layer = 18 ft

initial pressure = 0.75 ton/ft^2

final pressure = 1.5 ton/ft^2

Δp = 0.75 ton/ft^2

eo = 1.12

e1 = 0.98

k = 4.3 * 10^-7 cm/sec

A ) determine how long it will take to reach 70% consolidation

attached below is the detailed solution

T( time in days ) = 294.34 days

B) determine settlement at 70%

attached below is the detailed solution

Δh = 25.361 cm

4 0

2 years ago

1. Nitrogen at an initial state of 300 K, 150 kPa, and 0.2 m3 is compressed slowly in an isothermal process to a final pressure

Sauron [17]

Answer:

Workdone during the process = 130kJ

Explanation:

Workdone by an expanding gas which is also called pressure-volume work, is defined as the pressure exerted by the gas molecules on the walls of the containing vessel. If work done by an expanding gas is the energy transferred to its surroundings i.e If volume increases, workdone is negative(loses energy).

In an isothermal process(constant Temperature of 300K, we use boyles law:

P1V1 = P2V2

V2 = (150 * 0.2)/800

= 0.0375 m3

W = -P(V2 - V1)

= 800*(0.0375 - 0.200)

= 130kJ

7 0

3 years ago

A power cycle receives QH by heat transfer from a hot reservoir at TH = 1200 K and rejects energy QC by heat transfer to a cold

PIT_PIT [208]

Answer:

a) Irreversible, b) Reversible, c) Irreversible, d) Impossible.

Explanation:

Maximum theoretical efficiency for a power cycle ( $\eta_{r}$ ), no unit, is modelled after the Carnot Cycle, which represents a reversible thermodynamic process:

$\eta_{r} = \left(1-\frac{T_{C}}{T_{H}} \right)\times 100\,\%$ (1)

Where:

$T_{C}$ - Temperature of the cold reservoir, in Kelvin.

$T_{H}$ - Temperature of the hot reservoir, in Kelvin.

The maximum theoretical efficiency associated with this power cycle is: ( $T_{C} = 400\,K$ , $T_{H} = 1200\,K$ )

$\eta_{r} = \left(1-\frac{400\,K}{1200\,K} \right)\times 100\,\%$

$\eta_{r} = 66.667\,\%$

In exchange, real efficiency for a power cycle ( $\eta$ ), no unit, is defined by this expression:

$\eta = \left(1-\frac{Q_{C}}{Q_{H}}\right) \times 100\,\% = \left(\frac{W_{C}}{Q_{H}} \right)\times 100\,\% = \left(\frac{W_{C}}{Q_{C} + W_{C}} \right)\times 100\,\%$ (2)

Where:

$Q_{C}$ - Heat released to cold reservoir, in kilojoules.

$Q_{H}$ - Heat gained from hot reservoir, in kilojoules.

$W_{C}$ - Power generated within power cycle, in kilojoules.

A power cycle operates irreversibly for $\eta < \eta_{r}$ , reversibily for $\eta = \eta_{r}$ and it is impossible for $\eta > \eta_{r}$ .

Now we proceed to solve for each case:

a) $Q_{H} = 900\,kJ$ , $W_{C} = 450\,kJ$

$\eta = \left(\frac{450\,kJ}{900\,kJ} \right)\times 100\,\%$

$\eta = 50\,\%$

Since $\eta < \eta_{r}$ , the power cycle operates irreversibly.

b) $Q_{H} = 900\,kJ$ , $Q_{C} = 300\,kJ$

$\eta = \left(1-\frac{300\,kJ}{900\,kJ} \right)\times 100\,\%$

$\eta = 66.667\,\%$

Since $\eta = \eta_{r}$ , the power cycle operates reversibly.

c) $W_{C} = 600\,kJ$ , $Q_{C} = 400\,kJ$

$\eta = \left(\frac{600\,kJ}{600\,kJ + 400\,kJ} \right)\times 100\,\%$

$\eta = 60\,\%$

Since $\eta < \eta_{r}$ , the power cycle operates irreversibly.

d) Since $\eta > \eta_{r}$ , the power cycle is impossible.

8 0

2 years ago