Charge q is 1 unit of distance away from the source charge S. Charge p is two times further away. The force exerted between S and q is _____ the force exerted between S and p
a. 1/2
b. 2 times
c. 1/4
d. 4 times
Answer:
4 times
Step-by-step explanation:
Given the following :
Distance between q and S (r) = 1 unit
Distance between P and S (r) = 2 × 1 unit = 2
According to coliumbs law:
F = Ke(q1q2) /r^2
Where F = electrostatic force, Ke = coloumbs constant
q1,q2 = charges, r = distance between two charges
For q and S :
Fsq = Ke(q1q2) / 1^2
Fsq = Ke(q1q2) - - - - - equation 1
For P and S :
Fsp = Ke(q1q2) / 2^2
Fsp = Ke(q1q2) / 4 - - - - - equation 2
Dibiding equation 1 by equation 2
Fsq/Fsp = Ke(q1q2) / Ke(q1q2)/4
Fsq / Fsp = Ke(q1q2) × 4 / Ke(q1q2)
Fsq / Fsp = 4 / 1
Using cross multiplication
Therefore, Fsq = 4 × Fsp
132 because 11 times 4 times 3 is 132. Forget the 5 because the 11 and 5 is measuring basically the same thing.
H< 30
That is your inequality.
Answer:
C
Step-by-step explanation:
For the given intervals
( - ∞, - 5) ← use any value < - 5 but not - 5, the parenthesis ) indicates that x is less than - 5 but not equal to - 5
(- 5, - 1) ← - 4, - 3, - 2 can be used but not - 5 or - 1
(- 1, 4) ← 0, 1, 2, 3 can be used but not - 1 or 4
(4, ∞ ) ← use any value > 4 but not 4
Hence
3 can be used in (- 1, 4)
- 6 can be used in (- ∞, - 5)
zero can be used in (- 1, 4)
- 5 cannot be used in any of the given intervals
