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nirvana33 [79]
3 years ago
6

PLZ HURRY IM ON A TIMER

Engineering
1 answer:
Vinil7 [7]3 years ago
4 0

Answer:

he probably  works as an energy distributor  

Explanation:

he is controlling what gets put where

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2. BCD uses 6 bits to represent a symbol. a) True b) False​
Goshia [24]

Answer:

true because BCD used 6 bits to represent a symbol .

Explanation:

mark me brainlist

4 0
2 years ago
What does snow fall from?
Klio2033 [76]

Answer:

Clouds

Explanation:

It is created by trapped dust and water.

4 0
3 years ago
Use the graph to determine which statement is true about the end behavior of f(x).
Airida [17]

Answer:

As the x-values go to negative infinity, the function’s values go to positive infinity.

Explanation:

if the ans choices are:

As the x-values go to negative infinity, the function’s values go to negative infinity.

As the x-values go to negative infinity, the function’s values go to positive infinity.

As the x-values go to positive infinity, the function’s values go to negative infinity.

As the x-values go to positive infinity, the function’s values go to zero.

the ans is the 2nd choice

4 0
3 years ago
Read 2 more answers
Why is it better for a CPU to have more than one cache?
Tomtit [17]

Answer:

In general a cache memory is useful because the speed of the processor is higher than the speed of the ram . so reducing the number of memory is desirable to increase performance .

Explanation:

.

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#hope it helps you ..

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3 0
2 years ago
Steam enters a turbine in a Rankine cycle power plant at 200 psia and 500 °F. a) Calculate the isentropic thermal efficiency if
Aleks04 [339]

Answer:

η=0.19=19% for p=14.7psi

η=0.3=30% for p=1psi

Explanation:

enthalpy before the turbine, state: superheated steam

h1(p=200psi,t=500F)=2951.9KJ/kg

s1=6.8kJ/kgK

Entalpy after the turbine

h2(p=14.7psia, s=6.8)=2469KJ/Kg

Entalpy  before the boiler

h3=(p=14.7psia,x=0)=419KJ/Kg

Learn to pronounce

the efficiency for a simple rankine cycle is

η=\frac{h1-h2}{h1-h3}

η=(2951.9KJ/kg-2469KJ/Kg)/(2951.9KJ/kg-419KJ/Kg)

η=0.19=19%

second part

h2(p=1psia, s=6.8)=2110

h3(p=1psia, x=0)=162.1

η=(2951.9KJ/kg-2110KJ/Kg)/(2951.9KJ/kg-162.1KJ/Kg)

η=0.3=30%

7 0
3 years ago
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