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svlad2 [7]
2 years ago
10

1 . How are encoders used in the measurement of speed? Explain the encoder with a neat diagram.​

Engineering
1 answer:
rusak2 [61]2 years ago
4 0

The most common use for encoders is to measure angular or linear distance, but encoders can also be used to perform speed or velocity measurements. In other words, as the encoder rotates faster, the pulse frequency increases at the same rate

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(a)Compute the electrical conductivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in leng
igor_vitrenko [27]

Answer:

a) \sigma = 12.2 (Ω-m)^{-1}

b) Resistance = 121.4 Ω

Explanation:

given data:

diameter is 7.0 mm

length 57 mm

current I = 0.25 A

voltage v = 24 v

distance between the probes is 45 mm

electrical conductivity is given as

\sigma = \frac{I l}{V \pi r^2}

\sigma  = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}

\sigma = 12.2(Ω-m)^{-1}[/tex]

b)

Resistance = \frac{l}{\sigma A}

                  = \frac{l}{ \sigma \pi r^2}

= \frac{57  \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}

Resistance = 121.4 Ω

8 0
3 years ago
The contact angle between the mercury surface and capillary tube wall is______ A) Less than 90 B) Equal to 90 C) Greater than 90
MakcuM [25]

Answer:

The Answer to the question is :

Explanation:

The contact angle between the mercury surface and capillary tube wall is Greater than 90.

If the surface of the solid is hydrophobic, the contact angle will be greater than 90 °. On very hydrophobic surfaces the angle can be greater than 150º and even close to 180º.

8 0
3 years ago
What new information about hte role of the y subunit of ATP Synthase did this researcdh project elucidate? how does the subunit
nignag [31]

Answer:

Explanation:

it is observed given the information by the y subunit of ATP synthase that when one variable is changed, F1 motor becomes reversible and this is achieved by having the subunit as the reversible motor.

4 0
3 years ago
What is over head line
CaHeK987 [17]

Explanation:

An overhead power line is a structure used in electric power transmission and distribution to transmit electrical energy across large distances. It consists of one or more conductors (commonly multiples of three) suspended by towers or poles. Since most of the insulation is provided by air, overhead power lines are generally the lowest-cost method of power transmission for large quantities of electric energy.

<h3><em><u>Constr</u></em><em><u>uction</u></em></h3>

Towers for support of the lines are made of wood (as-grown or laminated), steel or aluminum (either lattice structures or tubular poles), concrete, and occasionally reinforced plastics. The bare wire conductors on the line are generally made of aluminum (either plain or reinforced with steel or composite materials such as carbon and glass fiber), though some copper wires are used in medium-voltage distribution and low-voltage connections to customer premises. A major goal of overhead power line design is to maintain adequate clearance between energized conductors and the ground so as to prevent dangerous contact with the line, and to provide reliable support for the conductors, resilience to storms, ice loads, earthquakes and other potential damage causes. Today overhead lines are routinely operated at voltages exceeding 765,000 volts between conductors.

<em>Please</em><em> </em><em>mark</em><em> </em><em>it</em><em> </em><em>as</em><em> </em><em><u>brainliest</u></em><em>. </em><em>Follow</em><em> </em><em>me</em><em> </em><em>I </em><em>w</em><em>ill</em><em> </em><em>fo</em><em>llow</em><em> you</em><em> back</em><em>. </em>

6 0
3 years ago
Two blocks A and B have a weight of 10 lb and 6 lb , respectively. They are resting on the incline for which the coefficients of
sertanlavr [38]

Answer:

Part A- 10.06^\circ

Part B- 0.1946ft

Explanation:

Block A

Apply force equilibrium equation along x direction.

\begin{array}{l}\\\sum {{F_x}} = 0\\\\{F_A} + {F_S} - 10\sin \theta = 0\\\end{array}

Here, {F_A} is friction force on block A and {F_S}is spring force, and \theta is inclination angle.

By substitution

0.14\,{N_A} + \left( {1.9x} \right)\,{\rm{lb}} \cdot {\rm{ft}} - 10\sin \theta = 0

Similarly, apply force equilibrium equation along y direction.

\begin{array}{l}\\\sum {{F_y}} = 0\\\\{N_A} - 10\cos \theta = 0\\\\{N_A} = 10\cos \theta \\\end{array}

By substitution

\begin{array}{l}\\0.14\,\left( {10\cos \theta } \right) + \left( {1.9x} \right) - 10\sin \theta = 0\\\\\left( {1.9x} \right) = 10\sin \theta - 1.4\cos \theta \\\end{array}

Block B

Apply force equilibrium equation along x direction.

\begin{array}{l}\\\sum {{F_x}} = 0\\\\{F_B} - {F_S} - 6\sin \theta = 0\\\end{array}

By substitution

0.24\,{N_A} - \left( {1.9x} \right) - 6\sin \theta = 0

Apply force equilibrium equation along y direction.

\begin{array}{l}\\\sum {{F_y}} = 0\\\\{N_B} - 6\cos \theta = 0\\\\{N_B} = 6\cos \theta \\\end{array}

\begin{array}{l}\\0.24\,\left( {6\cos \theta } \right) - \left( {1.9x} \right) - 6\sin \theta = 0\\\\\left( {1.9x} \right) = 1.44\cos \theta - 6\sin \theta \\\end{array}

Calculate the inclination angle when both blocks begin to slide.

\begin{array}{l}\\10\sin \theta - 1.4\cos \theta = 1.44\cos \theta - 6\sin \theta \\\\16\sin \theta = 2.84\cos \theta \\\\\tan \theta = \frac{{2.84}}{{16}}\\\\\theta = {\tan ^{ - 1}}\left( {\frac{{2.84}}{{16}}} \right)\\\end{array}

Calculate the change in length of the spring.

\begin{array}{l}\\\left( {1.9x} \right) = 1.44\cos \left( {10.06^\circ } \right) - 6\sin \left( {10.06^\circ } \right)\\\\1.9x = 0.3697\\\\x = 0.1946\,{\rm{ft}}\\\end{array}

3 0
3 years ago
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