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aev [14]
4 years ago
8

The input power for a thermostat is wired to the

Engineering
1 answer:
masya89 [10]4 years ago
6 0

Answer:

A. R

Explanation:

There are basically two wires that supply input power to the thermostat namely the C wire which is the common wire and the R wire .The G wire simply completes the input power circuit from the R-leg of the power supply.On the another hand the Y wire also completes the circuit  from the compressor fan contactor to the R-leg of the power supply . While the W wire completes the path to the heater contactor coil.

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16 . You are turning onto a two-lane road divided by a broken yellow line. You know immediately that:
Over [174]

When a person is turning onto a two-lane road divided by a broken yellow line, you know immediately that you are on a two-way road.

<h3>What is the road about?</h3>

Note that a Yellow centerlines can be seen in roads and it is one that is often used to separate traffic moving in different directions.

Note also that Broken lines can be crossed to allow slower-moving traffic and as such, When a person is turning onto a two-lane road divided by a broken yellow line, you know immediately that you are on a two-way road.

See full question below

You are turning onto a two-lane road divided by a broken yellow line. You know immediately that:

Answers

You are on a two-way road.

You are on a one-way road.

The road is under repair.

You must stay to the left of the broken yellow lines.

Learn more about  two-way road from

brainly.com/question/13123201

#SPJ2

5 0
2 years ago
To increase the thermal efficiency of a reversible power cycle operating between thermal reservoirs at TH and Tc, would you incr
alukav5142 [94]

<u></u>\ T_{c} has greater effect.

<u>Explanation</u>:

\eta_{\max }=1-\frac{T_{c}}{T_{A}}

T_{c}\\ = Temperature of cold reservoir

T_{H} = Temperature of hot reservoir

when T_{c} is decreased by 't',

$\eta_{\text {incre }}$ = 1-\frac{\left(\tau_{c}-t\right)}{T_{H}}

=n \ + \frac{t}{T_{n}}      -(i)

when {T_{H}} is increased by 'T'

\eta_{i n c}=\frac{n+\frac{t}{T_{H}}}{\left(1+\frac{k}{T_{H}}\right)}-(ii)

\eta_{\text {incre }} \ T_{c}>\eta_{\text {incre }} T_{\text {H }}

7 0
3 years ago
What the answer fast
anygoal [31]
Viscosity isT=u(U/y) where T is shear stress & u is velocity and y is thr length
The answer is =2.57
7 0
3 years ago
Pouring molten aluminum into a mold and allowing it to cool forms?
neonofarm [45]

Answer:it forms a molten mold that makes it hard to be able to smash something into it then make something like a key

Explanation:

7 0
3 years ago
Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you
viktelen [127]

Answer:

\theta_1=15^o\\\theta_2=75^o

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

V_x=V_{ox}=V_ocos\theta

Where \theta is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

V_y=V_{oy}-gt=V_osin\theta-gt

The  horizontal and vertical distances are, respectively:

x=V_{o}cos\theta t

\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

\displaystyle t_f=\frac{2V_osin\theta}{g}

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

\displaystyle R=\frac{V_o^2sin2\theta}{g}

Let's solve for \theta

\displaystyle sin2\theta=\frac{R.g}{V_o^2}

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}

\displaystyle \theta_2=\frac{180^o-asin\left(\frac{R.g}{V_o^2}\right)}{2}

Or equivalently:

\theta_2=90^o-\theta_1

Given Vo=37 m/s and R=70 m

\displaystyle \theta_1=\frac{asin\left(\frac{70\times 9.8}{37^2}\right)}{2}

\theta_1=15^o

And

\theta_2=90^o-15^o=75^o

5 0
3 years ago
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