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egoroff_w [7]
3 years ago
14

If you log into the admin account on windows 10, will the admin be notified ? ​

Engineering
2 answers:
professor190 [17]3 years ago
6 0

Answer:

Just message the Admin,

Ok.

tankabanditka [31]3 years ago
3 0
Most likely, sense now n days it says “you have been singled out” or “you have connected to somewhere else”
You might be interested in
3.
Andreyy89

Answer:

7

Explanation:

5 + 2 = 7

4 0
3 years ago
In normal operation, a paper mill generates excess steam at 20 bar and 400◦C. It is planned to use this steam as the feed to a t
Keith_Richards [23]

Answer:

The maximum power that can be generated is 127.788 kW

Explanation:

Using the steam table

Enthalpy at 20 bar = 2799 kJ/kg

Enthalpy at 2 bar = 2707 kJ/kg

Change in enthalpy = 2799 - 2707 = 92 kJ/kg

Mass flow rate of steam = 5000 kg/hr = 5000 kJ/hr × 1 hr/3600 s = 1.389 kg/s

Maximum power generated = change in enthalpy × mass flow rate = 92 kJ/kg × 1.389 kg/s = 127.788 kJ/s = 127.788 kW

6 0
3 years ago
Help now please evaluate using the commutative property: 40 (32) (10) (25)
Ilia_Sergeevich [38]

Answer:

#See solution for details.

Explanation:

#The commutative property of multiplication tells us that it doesn't matter in what order you multiply numbers. The formula for this property is a * b = b * a:

40\times32\times10\times 25=320,000\\\\25\times10\times32\times 40=320,000\\\\10\times32\times25\times 40=320,000\\\\32\times25\times40\times 10=320,000

Hence, the product of the four numbers remains 320,000 irrespective of their order.

4 0
3 years ago
Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp
telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

-0.40= 1.005(ln T_2 - 5.68697)- 0.5968

Solving for T2 we have:

T_2 = 5.8828

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

w_in = c_p(T_2 - T_1)+q_out

w_in = 1.005(358.8 - 295)+120

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = \frac{q_out}{T_1}

\frac{120kJ/kg.k}{295K}

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

3 0
3 years ago
An experimentalist claims that, based on his measurements, a heat engine receives 300 Btu of heat from a source of 900 R, conver
lesya [120]

Answer:

Hook's law holds good up to. A elastic limit. B. plastic limit. C.yield point. D.Breaking point

3 0
3 years ago
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