Diffrent ways to make a header above a window

A murder in a downtown office building has been widely publicized. You’re a police detective and receive a phone call from a dig

BaLLatris [955]

Answer:

Considering the plain view doctrine, which is an exception to the warrant requirement of the Fourth Amendment, is applied by law enforcement officers and courts who can seize evidence of a crime without a warrant, if the officer observes the evidence in plain view.

Explanation:

For any digital information related to a murder case that has been seized under the plain view doctrine to be used to convict you of a crime, has to comply with three conditions:

1. The digital evidence must be in out in the open, and easily observable by the officer, this is what "plain view" refers to.

2. The officer must have a legal right to be where he got the information related to the case.

3. The 'incriminating' character of the information must be a clear hint of the murder to fall under the plain view doctrine and the officer´s experience will help him determining whether the information is evidence or not, upon probable cause related to a crime.

7 0

3 years ago

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Calculate the density of the FCC nickel lattice with an interstitial hydrogen in the centered position of the unit cell. You may

Anastaziya [24]

Answer:

$\rho=8907.94\ Kg/m^3$

Explanation:

Given that

a=3.524 A

At.Wt. ,M= 58.7 g/mole,

For FCC

Z = 4

$4r=\sqrt2\ a$

The density given as

$\rho=\dfrac{ZM}{N_Aa^3}$

$\rho=\dfrac{4\times 58.7\times 10^{-3} }{ 6.023\times 10^{23}\times (3.524\times 10^{-10})^3}$

$\rho=8907.94\ Kg/m^3$

So the density is $\rho=8907.94\ Kg/m^3$

4 0

3 years ago

Phosphorus and nitrogen are included in which category of water pollutants?

Evgesh-ka [11]

Answer: Hello :)

They are in the <u>nutrient pollution</u> category.

Explanation:

3 0

2 years ago

An electrochemical cell is composed of pure nickel and pure iron electrodes immersed in solutions of their divalent ions. If the

xenn [34]

Answer:C 0.12 V

Explanation:

Given

Concentration of $Fe^{2+} M_1=0.40 M$

Concentration of $Ni^{2+} M_2=0.002 M$

Standard Potential for Ni and Fe are $V_2=-0.25 V$ and $V_1=-0.44 V$

$\Delta V=V_2-V_1-\frac{0.0592}{2}\log (\frac{M_1}{M_2})$

$\Delta V=-0.25-(-0.44)-\frac{0.0592}{2}\log (\frac{0.4}{0.002})$

$\Delta V=0.12\ V$

7 0

3 years ago

Air at 26 kPa, 230 K, and 220 rn/s enters a turbojet engine in flight. The air mass flow rate is 25 kg/s. The compressor pressur

Paha777 [63]

Answer:

Explanation:

Answer:

Explanation:

Answer:

Explanation:

This is a little lengthy and tricky, but nevertheless i would give a step by step analysis to make this as simple as possible.

(a). here we are asked to determine the Temperature and Pressure.

Given that the properties of Air;

ha = 230.02 KJ/Kg

Ta = 230 K

Pra = 0.5477

From the energy balance equation for a diffuser;

ha + Va²/2 = h₁ + V₁²/2

h₁ = ha + Va²/2 (where V₁²/2 = 0)

h₁ = 230.02 + 220²/2 ˣ 1/10³

h₁ = 254.22 KJ/Kg

⇒ now we obtain the properties of air at h₁ = 254.22 KJ/Kg

from this we have;

Pr₁ = 0.7329 + (0.8405 - 0.7329)[(254.22 - 250.05) / (260.09 - 250.05)]

Pr₁ = 0.77759

therefore T₁ = 254.15K

P₁ = (Pr₁/Pra)Pa

= 0.77759/0.5477 ˣ 26

P₁ = 36.91 kPa

now we calculate Pr₂

Pr₂ = Pr₁ (P₂/P₁) = 0.77759 ˣ 11 = 8.55349

⇒ now we obtain properties of air at

Pr₂ = 8.55349 and h₂ = 505.387 KJ/Kg

calculating the enthalpy of air at state 2

ηc = h₁ - h₂ / h₁ - h₂

0.85 = 254.22 - 505.387 / 254.22 - h₂

h₂ = 549.71 KJ/Kg

to obtain the properties of air at h₂ = 549.71 KJ/Kg

T₂ = 545.15 K

⇒ to calculate the pressure of air at state 2

P₂/P₁ = 11

P₂ = 11 ˣ 36.913

p₂ = 406.043 kPa

but pressure of air at state 3 is the same,

i.e. P₂ = P₃ = 406.043 kPa

P₃ = 406.043 kPa

To obtain the properties of air at

T₃ = 1400 K, h₃ = 1515.42 kJ/Kg and Pr = 450.5

for cases of turbojet engine,

we have that work output from turbine = work input to the compressor

Wt = Wr

(h₃ - h₄) = (h₂ - h₁)

h₄ = h₃ - h₂ + h₁

= 1515.42 - 549.71 + 254.22

h₄ = 1219.93 kJ/Kg

properties of air at h₄ = 1219.93 kJ/Kg

T₄ = 1140 + (1160 - 1140) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

T₄ = 1150.58 K

Pr₄ = 193.1 + (207.2 - 193.1) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

Pr₄ = 200.5636

Calculating the ideal enthalpy of the air at state 4;

Лr = h₃ - h₄ / h₃ - h₄*

0.9 = 1515.42 - 1219.93 / 1515.42 - h₄

h₄* = 1187.09 kJ/Kg

now to obtain the properties of air at h₄⁻ = 1187.09 kJ/Kg

P₄* = 179.7 + (193.1 - 179.7) [(1187.09 -1184.28) / (1207.57 - 1184.28)]

P₄* = 181.316

P₄ = (Pr₄/Pr₃)P₃ i.e. 3-4 isentropic process

P₄ = 181.316/450.5 * 406.043

P₄ = 163.42 kPa

For the 4-5 process;

Pr₅ = (P₅/P₄)Pr₄

Pr₅ = 26/163.42 * 200.56 = 31.9095

to obtain the properties of air at Pr₅ = 31.9095

h₅= 724.04 + (734.82 - 724.04) [(31.9095 - 3038) / (32.02 - 30.38)]

h₅ = 734.09 KJ/Kg

T₅ = 710 + (720 - 710) [(31.9095 - 3038) / (32.02 - 30.38)]

T₅ = 719.32 K

(b) Now we are asked to calculate the rate of heat addition to the air passing through the combustor;

QH = m(h₃-h₂)

QH = 25(1515.42 - 549.71)

QH = 24142.75 kW

(c). To calculate the velocity at the nozzle exit;

we apply steady energy equation of a flow to nozzle

h₄ + V₄²/2 = h₅ + V₅²/2

h₄ + 0 = h₅₅ + V₅²/2

1219.9 ˣ 10³ = 734.09 ˣ 10³ + V₅²/2

therefore, V₅ = 985.74 m/s

cheers i hope this helps

6 0

3 years ago