please answer all of them please Part A When two standard di
please answer all of them please
Part A When two standard dice are thrown, what is the probability that the sum of the dots on the two top faces will be 11? Express your answer using three significant digits. p= 10.305 Submit My Answers Give Up Incorrect, Try Again; 6 attempts remainingSolution
(22.62) .
Given q1 = -3 x10 -6 C
q2 = -5 x10 -6 C
q3 = 7 x10 -6 C
Coulomb\'s constant K = 8.99 x10 9 Nm 2/C 2
X component of force on q3 due to charge q1 is F = Kq3q1 / 4 2
Since distance between q3 and q1 along X axis is = 4 m
Substitue values you get , F = (8.99x10 9 )(7 x10 -6)(3x10 -6)/16
= 11.799 x10 -3 N
X component of force on q3 due to charge q2 is F \' = Kq3q1 / 3 2
Since distance between q3 and q2 along X axis is = 3 m
Substitue values you get , F \' = (8.99x10 9 )(7 x10 -6)(5x10 -6)/9
= 34.96 x10 -3 N
Both F and F \' are in same direction.
Therefore net force on q3 due to q1 and q2 along x direction = F+F \'
= 46.76 x10 -3 N
F x = 47 x10 -3 N
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Y component of force on q3 due to charge q1 is F = Kq3q1 / 3 2
Since distance between q3 and q1 along Y axis is = 3 m
Substitue values you get , F = (8.99x10 9 )(7 x10 -6)(3x10 -6)/9
= 20.97 x10 -3 N
Y component of force on q3 due to charge q2 is F \' = Kq3q2 / 4 2
Since distance between q3 and q2 along X axis is = 3 m
Substitue values you get , F \' = (8.99x10 9 )(7 x10 -6)(5x10 -6)/16
= 19.66 x10 -3 N
Both F and F \' are in opposite in direction.
Therefore net force on q3 due to q1 and q2 along Y direction = F-F \'
= 1.31 x10 -3 N along negative Y direction.
F y = -1.3 x10 -3 N

