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morpeh [17]
3 years ago
14

Kjhwe ,kenwif ujwfeowlwfwfwfw...

Engineering
2 answers:
Rashid [163]3 years ago
5 0
I’m so confused but ok
quester [9]3 years ago
4 0

Answer:

Unclear.

Explanation:

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With 64 KB of memory and 8 bits in each memory location, how wide should the address bus be to access all 64 KB of memory? (k =
marishachu [46]

Answer:

16-bit wide

Explanation:

In order to find the width of the address bus, we need first to know how many memory cells it is needed to address.

If the size memory is 64 KB, this means that the memory size, in bytes, is equal to the following quantity:

64 KB = 2⁶ * 2¹⁰ bytes = 2¹⁶ bytes.

In order to address this quantity of cell positions, the address bus must be able to address 2¹⁶ bytes, so it must have 16-bit wide.

3 0
3 years ago
What is a problem that technology can help solve that problem?
Maslowich
Seeing what the other side of the world is doing right now
8 0
3 years ago
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Technician A says independent shops are not affiliated with vehicle manufacturers, but it is easy for technicians who work in th
KatRina [158]

Answer:

b

Explanation:

i did it yeater dayajsbs

8 0
3 years ago
Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb
guajiro [1.7K]

Answer:

\eta_{turbine} = 0.603 = 60.3\%

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in  vapor state:

h₂ = h_{g\ at\ 125KPa} = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than s_g and greater than s_f at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88

Now, we will find h_{2s}(enthalpy at the outlet for the isentropic process):

h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg

Now, the isentropic efficiency of the turbine can be given as follows:

\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%

3 0
3 years ago
Two streams of air enter a control volume: stream 1 enters at a rate of 0.05 kg / s at 300 kPa and 380 K, while stream 2 enters
alex41 [277]

Answer:

0.08kg/s

Explanation:

For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.

The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.

 

finally you use the two previous equations to make a system and find the mass flows

I attached procedure

5 0
3 years ago
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