Answer:
Solution is given in the attachments,below.
Would be A 1012 N/C because The magnitude of the electric field at distance r from a point charge q is E=k
e
q/r
2
, so
E=
(5.11×10
−11
m)
2
(8.99×10
9
N.m
2
/C
2
)(1.60×10
−19
C)
=5.51×10
11
N/C∼10
1
2N/C
making (e) the best choice for this question.
The distance the spring stretches is the answer.
I hope this helps.
Have a nice day.
Answer:
64.2 m/s
Explanation:
We are given that
Speed ,v=38 m/s
We have to find the maximum speed when his car reach on flat ground.
Using dimensional analysis
If 35% acceleration reduced by F(res) at 38 m/s
Then, 100% acceleration can be reduced by F(res) at v' m/s
Substitute the values
Hence, the maximum speed when his car can reach on flat ground=64.2 m/s