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
Answer:
39,088?
Step-by-step explanation:
1.291
1.251
1.341
1.331
1.321
1.311
1.281
1.271
1.261
15/100*x=39
x=39*100/15
x= 260 is the total number of people who were asked
Answer:
x=6,-4
Step-by-step explanation:
Ok so we know the 2 angles are congruent so you need to set them equal to each other. x^2+5x=7x+24. Then get the x's together so you subtract by 7x to get x^2-2x=24. You would then get 24 on the other side so subtract by 24 to get x^2-2x-24. You then would need to factor the equation out. The factored form would be (x-6)(x+4). Then set it equal to 0 to get 6 and -4.
