Explanation:
The given data is as follows.
F = ![5.75 \times 10^{-16} N](https://tex.z-dn.net/?f=5.75%20%5Ctimes%2010%5E%7B-16%7D%20N)
q = ![1.6 \times 10^{-19} C](https://tex.z-dn.net/?f=1.6%20%5Ctimes%2010%5E%7B-19%7D%20C)
v = 385 m/s
= 0.876
Now, we will calculate the magnitude of magnetic field as follows.
B = ![\frac{F}{qv sin (\theta)}](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7Bqv%20sin%20%28%5Ctheta%29%7D)
= ![\frac{5.75 \times 10^{-16} N}{1.6 \times 10^{-19} C \times 385 m/s \times 0.876}](https://tex.z-dn.net/?f=%5Cfrac%7B5.75%20%5Ctimes%2010%5E%7B-16%7D%20N%7D%7B1.6%20%5Ctimes%2010%5E%7B-19%7D%20C%20%5Ctimes%20385%20m%2Fs%20%5Ctimes%200.876%7D)
=
T
= 10.65 T
Thus, we can conclude that magnitude of the magnetic field is 10.65 T.
Answer:
i can't click the answer bottom but the answer is "17th to 18th century" i hope this helps
Coronoid process of the ulna
Answer:
The answer is C.
Explanation:
I guessed and it was right
The answer is C. Final position minus initial position.