First of all, there are not <u>just</u> two applications that are solely applicable to the electron beam welding process. There are MANY.
Please visit out website at the URL below and you can click the "View Application" button under each listed Industry segment to view case studies of commonly EB welded applications.
https://www.ptreb.com/electron-beam-welding-applications
And for more general information on our welding process, we have an informational section you can peruse as well:
https://www.ptreb.com/electron-beam-welding-information
Good luck with your assignment- we are glad to hear they are teaching about EBW in high school!!!
Using the formula: E = kQ / d² where E is the electric field, Q is the test charge in coulomb, and d is the distance.
E = kQ / d²
k = 9 x 10^9 N-m²/C²
Q = 6.4 x 10^-5 C
d = 2.5 x 10^-2 m
Substituting the given values to the equation, we have:
E = (9 x 10^9)(6.4 x 10^-5) / (2.5 x 10^-2) ²
Electric field at the test charge is 921600000 N/C
Answer:
1 question is- I believe 4. Melting Question 2 is Density
Explanation:
THOUGHT ABOUT IT!!!
If they are both traveling with the same speed that means that they will reach other in the middle of the line initially between them. In other word, each will have to travel the same amount before they reach other.
Now you can calculate the time it takes for only one locomotive to travel half of the total distance between them, and that time is equal to the time you are looking for.
Use
t = S1/2 / v
where t-time, S-distance traveled , v-velocity
The answer is TRUE, I'm pretty sure.
