The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.
The given parameters;
- height of the waterfall, h = 0.432 m
- distance of the Salmon from the waterfall, s = 3.17 m
- angle of projection of the Salmon, = 30.8º
The time of motion to fall from 0.432 m is calculated as;

The minimum velocity of the Salmon jumping at the given angle is calculated as;

Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.
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The correct option is D.
The model developed by Ptolemy has a lot of inconsistency and during the middle age additional explanation was offered for the claims made by the model. The model was very complicated because it was based on erroneous assumptions.
Copernicus model was simpler and some of his claims were correct.<span />
Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)

q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.
The upward force exerted on the board by the support is 530.8 N.
<h3>Upward force exerted on the board by the support</h3>
The sum of the upward forces is equal to sum of downward forces;
total downward forces = 52.8 N + 206 N + 272 N = 530.8 N
downward force = upward force = 530.8 N
Thus, the upward force exerted on the board by the support is 530.8 N.
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<span>A. Rocket A will travel farther horizontally than rocket B.
This is because from the x axis, 40 m/s at 90 degrees travels directly vertical. 40 m/s at 70 degrees is slightly horizontal, so it will travel further horizontally.</span>