Why is California a good place for engineers to build suspension bridges?

Your study space does not need to be quiet as long as you can ignore any noise coming from the space true or false?

Makovka662 [10]

Answer:

False

Explanation:

When you're studying, you need to make sure that you can focus properly. This means that you shouldn't be hungry or too full and that you should be well-rested, in a quiet room with good lighting and no distractions. Noise is never good when you need to memorize something. Some people can partially ignore it as long as it isn't too loud, but it will begin to bother them eventually. That's why it's better to study in a quiet room.

3 0

3 years ago

Read 2 more answers

A carbon resistor has a resistance of 976 ohms at 0 degrees C. Determine its resistance at 89 degrees C

nignag [31]

Answer:

1028.1184 Ohms

Explanation:

<u>Given the following data;</u>

Initial resistance, Ro = 976 Ohms
Initial temperature, T1 = 0°C
Final temperature, T2 = 89°C

Assuming the temperature coefficient of resistance for carbon at 0°C is equal to 0.0006 per degree Celsius.

To find determine its new resistance, we would use the mathematical expression for linear resistivity;

$R_{89} = R_{0} + R_{0}(\alpha T)$

Substituting into the equation, we have;

$R_{89} = 976 + 976*(0.0006*89)$

$R_{89} = 976 + 976*(0.0534)$

$R_{89} = 976 + 52.1184$

$R_{89} = 1028.1184 \ Ohms$

5 0

3 years ago

Working with which of these systems requires a technician that has been certified in an EPA-approved course?

makvit [3.9K]

EPA Regulations provides a certified course for the technicians involved in the Air-conditioning system.

Answer: Option (b)

<u>Explanation:</u>

The EPA regulation has implemented an act called the "Clean Air Act" under the "section of 609".

This act provides some basic requirements for EPA Regulation such as follows;

Refrigerant: This unit must be approved by EPA Regulations before being implemented into the atmosphere.
Servicing: This system provides a certified course for technicians in service and also approve them with proper refrigerant equipment.
Reuse Refrigerants: The use of recycled refrigerants must be properly monitored before it comes in to serve.

6 0

4 years ago

Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is

Leona [35]

Answer:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

$dQ = dU - dW$

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) $\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW$

As per work definition:

$dW = F*dr$

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

$dW = PA*dx$

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

$dW = - P*dV$

So the third integral in equation (1) is:

$\int\limits^1_2 {- P} \, dV$

Considering the gas as ideal, the pressure can be calculated as $P = \frac{n*R*T}{V}$ , so:

$\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV$

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

$\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

Replacion this and solving equation (1) between state 1 and 2:

$\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}$

The internal energy depends only on the temperature of the gas, so there is no internal energy change $U_{2} - U_{1} = 0$ , so the heat exchanged to the system equals the work done by the system:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

4 0

4 years ago

A three-point flexure test is conducted on a cylindrical specimen of aluminum oxide. The specimen radius is 5.0 mm and the dista

kondaur [170]

Answer:

Detailed solution is given in the attached diagram

7 0

3 years ago

Why is California a good place for engineers to build suspension bridges?￼

Why is California a good place for engineers to build suspension bridges?