QUESTION:
WHAT IS THE MAGNITUDE OF THE MAGNETIC FIELD AT RIGHT ANGLES TO THE PROTON'S PATH?
ANSWER:
=<em><u>☑</u></em><em><u> </u></em><em><u>2</u></em><em><u>.</u></em><em><u>4</u></em><em><u> </u></em><em><u>T</u></em>
Answer:
=140 cm²
Explanation:
1
____ ×20x 14
2
=1
___ x base × height
is the formula have a gr8 day :)
2
The magnitude of the resultant force on the balloon is 374.13 N.
The given forces from the image;
- <em>Upward force = 514 N</em>
- <em>Downward force = 267 N</em>
- <em>Eastward force = 678 N</em>
- <em>Westward force = 397 N</em>
The net vertical force on the balloon is calculated as follows;

The net horizontal force on the balloon is calculated as follows;

The magnitude of the resultant force on the balloon is calculated as follows;

Thus, the magnitude of the resultant force on the balloon is 374.13 N.
Learn more here:brainly.com/question/4404327
An atom would be your answer, so B!
Answer:

Explanation:
Using the conservation of energy we have:

Let's solve it for v:

So the speed at the lowest point is 
Now, using the conservation of momentum we have:

Therefore the speed of the block after the collision is 
I hope it helps you!