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Volgvan
3 years ago
6

Transcription machinery assembles at _______________.

Engineering
1 answer:
vivado [14]3 years ago
3 0

Answer:

The base

Explanation:

You might be interested in
Carnot heat engine A operates between 20ºC and 520ºC. Carnot heat engine B operates between 20ºC and 820ºC. Which Carnot heat en
nikklg [1K]

Answer:

engine B is more efficient.

Explanation:

We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.

We also kn ow that the efficiency of Carnot cycle given as follows  

\eta =1-\dfrac{T_1}{T_2}

Here temperature should be in Kelvin.

For engine A

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{520+273}

\eta =0.63

For engine B

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{820+273}

\eta =0.73

So from above we can say that engine B is more efficient.

4 0
3 years ago
You want to find all files on your server that have either the SGID or SUID permission set. Which command should you use to obta
aalyn [17]

Answer:

For SGID you type this

$ find . -perm /4000

For SUID you type this

$ find . -perm /2000

Explanation:

Auxiliary file permissions, that are commonly referred to as “special permissions” in Linux are needed in order to easily find files which have SUID (Setuid) and SGID (Setgid) set.

After typing

$ find directory -perm /permissions

Then type the commands in the attachment below to obtain a list of these files with SGID and SUID.

3 0
3 years ago
What is the shape of the output signal on a rigexpert analyzer?
Gekata [30.6K]

Answer:

Output signal shape: square, from 0.1 to 230 MHz. Output power: -10 dBm (at a load of 50 Ohms).

Explanation:

8 0
2 years ago
Read 2 more answers
malott, m. e. (2003). paradox of organizational change: engineering organizations with behavioral systems analysis.
mylen [45]

Answer:

Paradox of Organizational Change: Engineering Organizations with Behavioral Systems Analysis. by. Maria E. Malott.

7 0
1 year ago
g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,
irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where   r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where   r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0
3 years ago
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