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Lina20 [59]
3 years ago
8

When selecting the appropriate gear for intended direction of travel, automatic transmission equipped vehicles should be placed

in _______ while manual transmission equipped vehicles should be placed in ______. a) Drive, first gear b) First gear, second gear c) First gear, drive d) Second gear, first gear
Engineering
1 answer:
True [87]3 years ago
6 0

Answer:

Automatic transmissions should be in Drive and Manual transmissions should be in first gear.

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A museum has three rooms, each with a motion sensor (m0, m1, and m2) that outputs 1 when motion is detected. At night, the only
nordsb [41]

Answer:

a) see attachment

b) A= m0m1+ m1m2+ m0m2

see attachment for K-map

c) see attachment

Explanation:

a) see attachment for truth table

b) see attachment for k-map

A= m0m1+ m1m2+ m0m2

c) see attachment for gate level circuit

4 0
3 years ago
2.5
spin [16.1K]
222 is the answer your welcome
6 0
3 years ago
A circular bar is 800mm in length and 32mm in diameter. The bar is made from a material with a modulus of elasticity E = 150 GPa
Yuki888 [10]

Answer:

For any material if ∈ is the axial strain then the lateral strain is given by -μ∈ is the lateral strain in the object

Where,

μ is the poisson's ratio of the material

The longitudinal strain is calculated as follows

\varepsilon _{axial}=\frac{\Delta length}{Length_{original}}\\\\\therefore \varepsilon _{axial}=\frac{0.7}{800}=8.75\times 10^{-4}

Thus the lateral strain becomes

\varepsilon _{lateral}=-\mu\varepsilon _{axial}\\\\\varepsilon _{lateral}=-0.27\times 8.75\times 10^{-4}=-2.36\times 10^{-4}

now by definition of lateral strain we have

\varepsilon _{lateral}=\frac{\Delta diameter}{diameter_{original}}\\\\\Rightarrow \Delta Diameter=-2.36\times 10^{-4}\times 32=-7.56\times 10^{-3}\\\\D_{f}-D_{i}=-7.56\times 10^{-3}\\\\D_{f}=32-7.56\times 10^{-3}=31.992mm

By hookes law the stress developed due to the given strain is given by

\sigma =\varepsilon _{axial}E

Applying values we get

\sigma =8.75\times 10^{-4}\times 150\times 10^{9}\\\\\sigma =131.25MPa

Thus the force is calculated as

Force=\sigma \times Area\\\\Force=131.25\times 10^{6}\times \frac{\pi (32\times 10^{-3})^{4}}{4}\\\\Force=105.55kN

5 0
3 years ago
What is pre-flush and post flush in petroleum engineering?
cluponka [151]

Answer:

Pre-Flush:

It is also known as In-line Equalization. In this stage of flow equalization, all the flow passes through the equalization basin. It helps in reduction of fluctuation in pollutants concentration and flow rate and helps to control short term surges with the use of basin.

Post-Flush:

Another name for this stage is Off-line Equalization. In this stage, only overflow above a predetermined standard is diverted into the basin. It helps in reducing the fluctuations in loading by a considerable amount and helps to reduce the pumping requirement. It is basically used to capture "first flush" from combined collection systems.

5 0
3 years ago
Different metabolic control systems have different characteristic time scales for a control response to be achieved. Match the t
prisoha [69]

Answer:

a. Covalent modification = Seconds to minutes

b. Allosteric control = Milliseconds

c. Gene expression = Hours

Explanation:

Covalent modifications refer to the addition and/or removal of chemical groups by the action of particular enzymes such as methylases, acetylases, phosphorylases, phosphatases, etc. For example, histones are chromatin-associated proteins covalently modified by enzymes that add methyl groups (histone methylation), acetyl groups (histone acetylation), phosphate groups (histone phosphorylation), etc. Moreover, allosteric control, also known as allosteric regulation, is a type of regulation of the enzyme activity by binding an effector molecule (allosteric modulator) at a different site than the enzyme's active site, thereby triggering a conformational change on the enzyme upon binding of an effector. Finally, gene expression encompasses the cellular processes by which genetic information flows from genes to proteins (i.e., transcription >> translation). In metabolic pathways, enzymes that are able to catalyze irreversible reactions represent sites of control (for example, during glycolysis, pyruvate kinase is an enzyme that catalyzes an irreversible reaction, thereby serving as a control site). In turn, enzymatic activity is modulated by covalent modifications or reversible binding of allosteric effectors. Finally, metabolic pathways are also modulated by gene regulatory mechanisms that control the transcription of specific enzymes required for such pathways. During these processes, the times required for allosteric regulation, covalent modification (e.g., phosphorylation) and transcriptional control can be counted in milliseconds, seconds, and hours, respectively.

5 0
3 years ago
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