Answer:
Approximately
(given that the magnitude of this charge is
.)
Explanation:
If a charge of magnitude
is placed in an electric field of magnitude
, the magnitude of the electrostatic force on that charge would be
.
The magnitude of this charge is
. Apply the unit conversion
:
.
An electric field of magnitude
would exert on this charge a force with a magnitude of:
.
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.
Answer: 20
Explanation: Mass number is the number of neutrons plus the number of protons. 8 + 12 gives 20.
Answer:
For vector u, x component = 10.558 and y component =12.808
unit vector = 0.636 i+ 0.7716 j
For vector v, x component = 23.6316 and y component = -6.464
unit vector = 0.9645 i-0.2638 j
Explanation:
Let the vector u has magnitude 16.6
u makes an angle of 50.5° from x axis
So 
Vertical component 
So vector u will be u = 10.558 i+12.808 j
Unit vector 
Now in second case let vector v has a magnitude of 24.5
Making an angle with -15.3° from x axis
So horizontal component 
Vertical component 
So vector v will be 23.6316 i - 6.464 j
Unit vector of v 
Answer: I think it’s 20cm.
My educated guess : 21.2 deg