Good morning.
We have:

Where
j is the unitary vector in the direction of the
y-axis.
We have that

We add the vector
-a to both sides:

Therefore, the magnitude of
b is
47 units.
Well, since you only want direction, ignore the numbers. Use the right hand rule.
Current (pointer finger) points west (left).
Magnetic field (middle finger) points south (towards you).
Force (thumb) then points up (away from the earth)
Answer:
The magnitude is "3.8 m/s²", in the upward direction.
Explanation:
The given values are:
Mass,
m = 88 kg
Scale reads,
T = 900 N
As we know,
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
Now,
⇒ 
On substituting the given values in the above equation, we get
⇒ 
On subtracting "862.4" from both sides, we get
⇒ 
⇒ 
⇒ 
⇒
(upward direction)
Answer:
The net Electric field at the mid point is 289.19 N/C
Given:
Q = + 71 nC = 
Q' = + 42 nC = 
Separation distance, d = 1.9 m
Solution:
To find the magnitude of electric field at the mid point,
Electric field at the mid-point due to charge Q is given by:



Now,
Electric field at the mid-point due to charge Q' is given by:



Now,
The net Electric field is given by:


The answer is C. It would look similar to the graph for KNO3