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Katarina [22]
3 years ago
7

Consider the direction field of the differential equation dy/dx = x(y − 6)2 − 4, but do not use technology to obtain it. Describ

e the slopes of the lineal elements on the lines x = 0, y = 5, y = 6, and y = 7.
Physics
1 answer:
Alona [7]3 years ago
5 0

Answer:

The slopes shows that the direction of the field is from -2 to +2, with three point charges, q₁, q₂ and q₃ at -2, 0 and +2 respectively.

Explanation:

Given;

The slope, dy/dx = 2x(y-6) - 4

2x(y-6) - 4 = 2xy - 12x - 4, divide through by 'x'

dy/dx = 2y -12 - 4/x

The slopes of the linear elements on the lines, x =0, y = 5, y = 6, y = 7.

At x = 0, and y = 5

dy/dx = 2y -12 - 4/x

dy/dx = 2(5) - 12 = -2

At x = 0, and y = 6

dy/dx = 2y -12 - 4/x

dy/dx = 2(6) - 12 = 0

At x = 0, and y = 7

= 2y -12 - 4/x

dy/dx = 2(7) - 12 = 2

Therefore, the slopes shows that the direction of the field is from -2 to +2, with three point charges, q₁, q₂ and q₃ at -2, 0 and +2 respectively.

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The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

The given parameters;

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The vertical pressure of the water is calculated as follows;

P = \rho gh\\\\\frac{P}{h} = \rho g\\\\\frac{P}{h} = k\\\\\frac{P_1}{h_1} = \frac{P_2}{h_2} \\\\

The vertical height of the first floor from the 13th floor = 130 ft

The vertical height of the 13 ft floor = 10  ft

P_1 = \frac{P_2 h_1}{h_2} \\\\P_1 = \frac{35 \times 130}{10} \\\\P_1 = 455 \ PSI

Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

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