To determine the answer of Part A draw the equilateral triangle and the to determine the coordinates of of the third charge use that triangle.
To calculate the gravitational field strength in part B from each of the charges use the following equation.
E=kcq/r2
If you would add those values then you can use the symmetry about the y axis to make the vector addition a litter easier.<span />
Answer:
- as written, c ≈ 0.000979 or c = 4
- alternate interpretation: c = 0
Step-by-step explanation:
<em>As written</em>, you have an equation that cannot be solved algebraically.
(32^2)c = 8^c
1024c = 8^c
1024c -8^c = 0 . . . . . . rewrite as an expression compared to zero
A graphical solution shows two values for c: {0.000978551672551, 4}. We presume you're interested in c = 4.
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If you mean ...
32^(2c) = 8^c
(2^5)^(2c) = (2^3)^c . . . . rewriting as powers of 2
2^(10c) = 2^(3c) . . . . . . . simplify
10c = 3c . . . . . . . . . . . . . .log base 2
7c = 0 . . . . . . . . . . . . . . . subtract 3c
c = 0 . . . . . . . . . . . . . . . . divide by 7
Answer: z = 7
Step-by-step explanation:
Step 1: Subtract 4 from both sides
z + 4 - 4 = 11 -4
z = 7
Answer:
3.33 and 1/3
Step-by-step explanation:
"Dense" here means that there are infinite irrational numbers between two rational numbers. Also, there are infinite rational numbers between two rational numbers. That's the meaning of dense. Actually, that can be apply to all real numbers, there always is gonna be a number between other two.
But, to demonstrate that irrationals are dense, we have to based on an interval with rational limits, because the theorem about dense sets is about rationals, and the dense irrational set is a deduction from it. That's why the best option is 2, because that's an interval with rational limits.
