Answer:
5.25 m
Explanation:
Given;
The height equation h;
h=-x^2+3x+3
Where;
h = the height above water
x = horizontal distance from the end of the board
The maximum height is at h' = 0, when change in h with respect to change in x is equal to zero.
differentiating the equation h.
dh/dx = h' = -2x + 3 = 0
Solving for x;
2x = 3
x = 3/2
Substituting into the function h;
h max = -x^2+3x+3
h max = -(3/2)^2 + 3(3/2) +3 = -9/4 +9/2 +3 = 9/4 + 3 =
h max = 21/4 = 5.25 m
it's how much it weighs and how much force is pushing on it like a egg if i drop it the weigh can cause it to break and how much force the gravity is pushing on it.
We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2
F(1) = F (2on1) + F (3on1)
F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N
Since F1 is 7N,
F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)
Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N
F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m
So it's located in -.144m
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The answer to the question is True
