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goblinko [34]
3 years ago
9

What type of OS is very fast and small, and is often embedded, or built into the circuitry of a device?

Physics
1 answer:
Bezzdna [24]3 years ago
7 0

Answer:

Real time operating system (RTOS)

Explanation:

  • Real time operating system is an operating system software which is used for the real-time applications that processes the data as soon as the input comes to the system.

It is very fast because it processes the data without any buffer delays.

  • In RTOS the scheduler provides a deterministic time which is predicted for the execution pattern. In a real-time operation the system must respond to each sequence of execution within a certain time frame which is the deadline for the operation so that the next inline process can be executed. Hence it is used in embedded systems and circuitry devices.
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A 103 kg horizontal platform is a uniform disk of radius 1.71 m and can rotate about the vertical axis through its center. A 68.
Andreyy89

Answer:

I_{total}=220.64 kg*m^{2}

Explanation:

The moment of inertia of the system is equal to the each population and the platform inertia so

Inertia disk

I_{disk}=\frac{1}{2}*m_{disk}*(r_{p})^{2}

Inertia person

I_{p}=\frac{1}{2}*m_{p}*(r_{p})^{2}

Inertia dog

I_{d}=\frac{1}{2}*m_{d}*(r_{d})^{2}

The Inertia of the system is the sum of each mass taking into account that all exert the force of inertia:

I_{total}=I_{disk}+I_{p}+I_{d}

I_{total}=\frac{1}{2}*103kg*(1.71)^{2}+\frac{1}{2}*68.9kg*(1.09)^{2}+\frac{1}{2}*27.7kg*(1.45)^{2}

I_{total}=220.64 kg*m^{2}

5 0
3 years ago
During which stage of the cell cycle does the cell condense chromosomes and break down the nuclear envelope? Group of answer cho
Natalka [10]

Answer:

Prophase

Explanation:

During the Prophase

6 0
3 years ago
A water pump pumps 3 kg of water every second from the bottom of a well to the top of the well 50 m above if the pump used 2000
vesna_86 [32]

watts = work per second.

work is mgh = 3x10x50=1500

watts out = 1500

Watts used = 2000

eff=1500/2000=75%


8 0
3 years ago
A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C
Vinil7 [7]

Answer:

1)

75 kmh⁻¹

2)

75 kmh⁻¹

Explanation:

1)

v_{ab} = Speed of train from station A to station B = 80 kmh⁻¹

d_{ab} = distance traveled from station A to station B

t_{ab} = time of travel between station A to station B

we know that

Time = \frac{distance}{speed}

t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}

d_{bc} = distance traveled from station B to station C

v_{bc} = Speed of train from station B to station C = 60 kmh⁻¹

t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}

Total distance traveled is given as

d = d_{ab} + d_{bc}

Total time of travel is given as

t = t_{ab} + t_{bc}

Average speed is given as

v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }

Given that :

d_{ab} = 4 d_{bc}

So

v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}

2)

v_{ab} = Speed of train from station A to station B = 80 kmh⁻¹

t_{ab} = time of travel between station A to station B

d_{ab} = distance traveled from station A to station B

we know that

distance = (speed) (time)

d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}

d_{bc} = distance traveled from station B to station C

v_{bc} = Speed of train from station B to station C = 60 kmh⁻¹

t_{bc} = time of travel for train from station B to station C

we know that

distance = (speed) (time)

d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}

Total distance traveled is given as

d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}

Total time of travel is given as

t = t_{ab} + t_{bc}

Average speed is given as

v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}

Given that :

t_{ab} = 3 t_{bc}

So

v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}

4 0
3 years ago
What’s the answer???
inna [77]

Answer: ( 2nd ) ( 3 )

Explanation:

4 0
2 years ago
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