Question 2 (Multiple Choice Worth 3 points)

Consider a standard room thermostat. Determine the sensor, transducer, output, and control stages for this measurement system.

Anni [7]

Answer:

Sensor/transducer: bimetallic thermometer

Output: displacement of thermometer tip

Control Tstages: mercury contact switch (open:furnace off; closed:furnace on

Explanation:

for a standard room thermostat : This is the device that sets/determines the temperature of an enclosure.

Sensor/transducer: bimetallic thermometer: Bimetalic thermometer are used for measuring the temperature of the ambient air . bimetallic thermometer actually contains two metals. they undergo linear expansivity as the temperature of the room changes.in other words, they experience contraction and expansion with increase or decrease in temperature.The sensor is basically coupled with a transducer which turns the measured variable(Temperature) into something else, such as a movement on a dial or an electrical signal

Output: displacement of thermometer tip

Controller: mercury contact switch (open:furnace off; closed:furnace on)

once the contact switch is open the furnace can go off. when the contact switch is closed, the furnace will come up.

3 0

3 years ago

On a average work day more than work place firs are reorted

Basile [38]

What is the question? It looks like a statement...

8 0

3 years ago

Anna makes arrangements to reuse waste water that has been used in sinks and showers. Which term refers to the waste water that

victus00 [196]

Answer:

Greywater.

Explanation:

Greywater is also known as sullage and it can be defined as any form of gently used wastewater derived from sources within a residential or office building such as showers, washing machines, bathroom sinks, bathroom tub, etc.

Generally, greywater or sullage is completely free of fecal materials (faeces) because it is independent from all toilet activities. However, greywater is not clean for direct use because it usually contains food particles, dirt, oil from dishes, hair, etc.

In this scenario, Anna makes arrangements to reuse waste water that has been used in sinks and showers. Greywater is a term which refers to the waste water that Anna reuses to conserve resources.

Therefore, Anna reuses greywater to conserve resources.

8 0

3 years ago

Which measuring tool will be used to determine the diameter of a crankshaft journal?

guapka [62]

Answer:

The dial bore gauge measures the inside of round holes, such as the bearing journals. This one tool can measure 2″ up to 6″ diameter holes. Both tools are needed in order to check the interior and exterior dimensions of the crankshaft, rods and engine block journals, as well as the thickness of the bearings themselves.

Hope it's helpful to you

pls mark me as brain list

4 0

3 years ago

While playing a game of catch on the quadrangle, you throw a ball at an initial velocity of 17.6 m/s (approximately 39.4 mi/hr),

MAXImum [283]

Answer:

a) The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) The ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

Explanation:

a) The ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical motion at constant acceleration. First, we calculate the time taken by the ball to hit the ground:

$y = y_{o} + (v_{o}\cdot \sin \theta) \cdot t+\frac{1}{2}\cdot g\cdot t^{2}$ (1)

Where:

$y_{o}$ , $y$ - Initial and final vertical position, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$\theta$ - Launch angle, measured in sexagesimal degrees.

$g$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $y_{o} = 2\,m$ , $y = 0\,m$ , $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $g = -9.807\,\frac{m}{s^{2}}$ , then the time taken by the ball is:

$-4.904\cdot t^{2}+13.482\cdot t +2 = 0$ (2)

This second order polynomial can be solved by Quadratic Formula:

$t_{1} \approx 2.890\,s$ and $t_{2} \approx -0.141\,s$

Only the first root offers a solution that is physically reasonable. That is, $t \approx 2.890\,s$ .

The vertical velocity of the ball is calculated by this expression:

$v_{y} = v_{o}\cdot \sin \theta +g\cdot t$ (3)

Where:

$v_{o,y}$ , $v_{y}$ - Initial and final vertical velocity, measured in meters per second.

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $t \approx 2.890\,s$ , then the final vertical velocity is:

$v_{y} = -14.860\,\frac{m}{s}$

The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) From a) we understand that ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball ( $x$ ) is determined by the following expression:

$x = (v_{o}\cdot \cos \theta)\cdot t$ (4)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $t \approx 2.890\,s$ , then the distance covered by the ball is:

$x = 32.695\,m$

The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before hitting the ground ( $v$ ), measured in meters per second, is determined by the following Pythagorean identity:

$v = \sqrt{(v_{o}\cdot \cos \theta )^{2}+v_{y}^{2}}$ (5)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , then the magnitude of the velocity of the ball is:

$v \approx 18.676\,\frac{m}{s}$ .

The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is defined by the following trigonometric relationship:

$\tan \theta = \frac{v_{y}}{v_{o}\cdot \cos \theta_{o}}$

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta_{o} = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , the angle of the total velocity of the ball just before hitting the ground is:

$\theta \approx -52.717^{\circ}$

The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

3 0

3 years ago

Read 2 more answers