A car is built from various subsystems. If these subsystems are not working properly it is dangerous because it can cause a serious traffic accident.
<h3>What subsystems do cars have?</h3>
When you're testing the build of a car, you have to check its many subsystems:
- the battery
- the engine
- the cabin
- the thermal-management system
- the gearbox
- the chassis
- the suspension
<h3>Why is a car with damaged subsystems dangerous?</h3>
The subsystems of a car are very important components that allow the proper functioning of the car. These subsystems work synchronously making the car work properly.
However, if one of these subsystems is not working properly it could cause a malfunction that could lead to a traffic accident.
Learn more about cars in: brainly.com/question/11733094
The average speed of the whole travel is equal to <u>400 mph</u>.
Why?
From the statement, we know that whole travel is divided into three parts. For the first part (traveling from New York to Chicago), we have that it was 3.25 hours and the covered distance was half of the total distance (1400mi). For the second part, we have that it was 1 hour (layover time), and the covered no distance. For the third part (traveling from Chicago to Los Angeles), we have that it was 2.75 hours, and it took the other half of the total distance (1400mi).
We can calculate the average speed of the whol travel using the following formula:
Now, substituting and calculating, we have:
Hence, we have the average speed of the whole travel is equal to 400 mph.
Have a nice day!
Answer:the witch has nothing to do with the problem
Explanation:
Answer:
The maximum energy stored in the combination is 0.0466Joules
Explanation:
The question is incomplete. Here is the complete question.
Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Energy stored in a capacitor is expressed as E = 1/2CtV² where
Ct is the total effective capacitance
V is the supply voltage
Since the capacitors are connected in series.
1/Ct = 1/C1+1/C2+1/C3
Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF
1/Ct = 1/11.7 + 1/21.0 + 1/28.8
1/Ct = 0.0855+0.0476+0.0347
1/Ct = 0.1678
Ct = 1/0.1678
Ct = 5.96μF
Ct = 5.96×10^-6F
Since V = 125V
E = 1/2(5.96×10^-6)(125)²
E = 0.0466Joules
