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VARVARA [1.3K]
2 years ago
7

It is ___ for motorcyclists to ride more than two abreast in a lane.

Engineering
1 answer:
nevsk [136]2 years ago
6 0
Legal! Hope this helps
You might be interested in
How to design a solar panel<br>​
artcher [175]

Answer:

#1) Find out how much power you need

#2 Calculate the amount of batteries you need.

#3 Calculate the number of solar panels needed for your location and time of year.

#4 Select a solar charge controller.

#5 Select an inverter.

#6 Balance of system

Explanation: To design solar panel, consider the following steps

1.) Find the power consumption demands

The first step in designing a solar PV system is to find out the total power and energy consumption of all loads that need to be supplied by the solar PV system as follows:

Calculate total Watt-hours per day for each appliance used.

 Add the Watt-hours needed for all appliances together to get the total Watt-hours per day which must be delivered to the appliances.

Calculate total Watt-hours per day needed from the PV modules.

Multiply the total appliances Watt-hours per day times 1.3 (the energy lost in the system) to get the total Watt-hours per day which must be provided by the panels.

2. Size the PV modules

Different size of PV modules will produce different amount of power. To find out the sizing of PV module, the total peak watt produced needs. The peak watt (Wp) produced depends on size of the PV module and climate of site location. We have to consider panel generation factor which is different in each site location. For Thailand, the panel generation factor is 3.43. To determine the sizing of PV modules, calculate as follows:

2.1 Calculate the total Watt-peak rating needed for PV modules

Divide the total Watt-hours per day needed from the PV modules (from item 1.2) by 3.43 to get the total Watt-peak rating needed for the PV panels needed to operate the appliances.

Calculate the number of PV panels for the system

Divide the answer obtained in item 2.1 by the rated output Watt-peak of the PV modules available to you. Increase any fractional part of result to the next highest full number and that will be the 

number of PV modules required.

Result of the calculation is the minimum number of PV panels. If more PV modules are installed, the system will perform better and battery life will be improved. If fewer PV modules are used, the system may not work at all during cloudy periods and battery life will be shortened.

3. Inverter sizing

An inverter is used in the system where AC power output is needed. The input rating of the inverter should never be lower than the total watt of appliances. The inverter must have the same nominal voltage as your battery.

For stand-alone systems, the inverter must be large enough to handle the total amount of Watts you will be using at one time. The inverter size should be 25-30% bigger than total Watts of appliances. In case of appliance type is motor or compressor then inverter size should be minimum 3 times the capacity of those appliances and must be added to the inverter capacity to handle surge current during starting.

For grid tie systems or grid connected systems, the input rating of the inverter should be same as PV array rating to allow for safe and efficient operation.

4. Battery sizing

The battery type recommended for using in solar PV system is deep cycle battery. Deep cycle battery is specifically designed for to be discharged to low energy level and rapid recharged or cycle charged and discharged day after day for years. The battery should be large enough to store sufficient energy to operate the appliances at night and cloudy days. To find out the size of battery, calculate as follows:

     4.1 Calculate total Watt-hours per day used by appliances.

     4.2 Divide the total Watt-hours per day used by 0.85 for battery loss.

     4.3 Divide the answer obtained in item 4.2 by 0.6 for depth of discharge.

     4.4 Divide the answer obtained in item 4.3 by the nominal battery voltage.

     4.5 Multiply the answer obtained in item 4.4 with days of autonomy (the number of days that you need the system to operate when there is no power produced by PV panels) to get the required Ampere-hour capacity of deep-cycle battery.

Battery Capacity (Ah) = Total Watt-hours per day used by appliancesx Days of autonomy

(0.85 x 0.6 x nominal battery voltage)

5. Solar charge controller sizing

The solar charge controller is typically rated against Amperage and Voltage capacities. Select the solar charge controller to match the voltage of PV array and batteries and then identify which type of solar charge controller is right for your application. Make sure that solar charge controller has enough capacity to handle the current from PV array.

For the series charge controller type, the sizing of controller depends on the total PV input current which is delivered to the controller and also depends on PV panel configuration (series or parallel configuration).

According to standard practice, the sizing of solar charge controller is to take the short circuit current (Isc) of the PV array, and multiply it by 1.3

Solar charge controller rating = Total short circuit current of PV array x 1.3

5 0
3 years ago
The state of plane strain on an element is:
balu736 [363]

Answer:

a. ε₁=-0.000317

   ε₂=0.000017

θ₁= -13.28° and  θ₂=76.72°  

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2}  \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2}  \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6}  \pm 1.67 \times 10^{-4}

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5

θ= -13.28° and  76.72°

To determine the direction of ε₁ and ε₂

\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2}  + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta  + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2}  + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56)  + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and  θ₂=76.72°  

b. maximum in-plane shear strain

\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}

=3.335 *10^-4

\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )

ε(avg) =150 *10^-6

orientation of γmax

tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}

θ = 31.71 or -58.29

To determine the direction of γmax

\gamma _{x'y' }=  - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta  + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }=  - \frac{-300*10^{-6} - \ 0}{2} sin(63.42)  + \frac{150*10^{-6}}{2}cos(63.42)

= 1.67 *10^-4

4 0
3 years ago
At what times should you use your headlights?
SVETLANKA909090 [29]

Answer:

Headlights are required to be used 1/2 hour after sunset to 1/2 hour before sunrise, when windshield wipers are being used, when visibility is less than 1000 feet, or when there is insufficient light or adverse weather.

Explanation:

hope this helps

8 0
3 years ago
Read 2 more answers
An engineer lives in Hawaii at a location where the annual rain fall is 300 inches. She decides to use the rain to generate elec
Alex777 [14]

Answer:

80.7lbft/hr

Explanation:

Flow rate of water in the system = 3.6x10^-6

The height h = 100

1s = 1/3600h

This implies that

Q = 3.6x10^-6/[1/3600]

Q = 0.0000036/0.000278

Q = 0.01295

Then the power is given as

P = rQh

The specific weight of water = 62.3 lb/ft³

P = 62.3 x 0.01295 x 100

P = 80.675lbft/h

When approximated

P = 80.7 lbft/h

This is the average power that could be generated in a year.

This answers the question and also corresponds with the answer in the question.

4 0
3 years ago
Refer to Figure 9-18. A #_____ electrode lead and workpiece lead should be used to carry 150 amperes of electricity 75 feet to t
Anni [7]

Answer:

AC or

Explanation:

BC

8 0
2 years ago
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