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mylen [45]
1 year ago
15

What can you say about a solution of the equation y' = (1/2)y2 just by looking at the differential equation?

Physics
1 answer:
noname [10]1 year ago
6 0

The function y must be equal to 0 on any interval on which it is defined. The function y must be increasing (or equal to 0) on any interval on which it is defined.

Analysis of solution by seeing differential equation:

Given differential equation is: y' = (1/2)y2

How do deduce the results just by seeing them?

The equation tells us that:

rate = positive of ( y^2 )

rate = positive of (positive or zero) = positive or zero

Thus, the rate is positive or zero no matter what value we put in the place of y from its valid domain, since.

When the rate is positive or zero, that means the function will never grow upwards. Thus, either increasing or staying at the same level.

Learn more about differential equations here:

brainly.com/question/25731911

#SPJ4

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Please help me with this question. BRAINLIEST if answered.
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... The top branch of the 3-branched parallel block ... the 9 and 6 in series ...
is equivalent to a single resistor of 15 ohms.

... The 3-branched parallel block boils down to (30, 10, and 15) in parallel.
That's (1/30 + 1/10 + 1/15)⁻¹  =   5 ohms.

... The 5-ohm-equivalent block and the 20-ohm resistor form a
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The voltage across the 5-ohm-equivalent block is (5/25 x 30v) = 6v .

... The top branch of the block is equivalent to a (9 + 6) = 15-ohmer.
With 6v across its ends, the current through that branch is (6/15) = 0.4A .

... With 0.4A flowing through it, the 9-ohm resistor is dissipating

            I²R = (0.4A)² (9 ohms)  =  (0.16 A²) (9 ohms)  =  1.44 W  (choice-3)   
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Water is traveling into what sphere during infiltration?
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3 years ago
Water enters a student's house 10.0 m above the ground through a pipe with a cross section area of 1.00 x 10-4m2 at ground. Insi
dezoksy [38]

Answer:

(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)

(b). P₂ = 236500 Pa

Explanation:

This is quite straight-forward so let us begin by defining the terms given.

Given that;

The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.

Let us make the velocity of water inside the house be V₁

such that the Volume of water entering the per second is = A₁V₁

Therefore, in 90sec:

45 L =  90 A₁V₁

V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴

V₁ = 10m/s            (velocity of water inside the house)

From the continuity equation we have that;

A₁V₁ = A₂V₂

0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂

V₂ = 5m/s               (velocity at ground level)

(b). We are told to calculate the water pressure in the pipeline at the ground level.

Using Bernoulli's equation;

P₁ + pgh₁ + 1/2PV₁²  (inside)      =       P₂ + pgh₂ + 1/2PV₂²   (ground level)

1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²

P₂ (pressure) = 1.01*10⁵Pa

Therefore we have;

101000 + 98000 + 50000 = P₂ + 12500

P₂ = 236500 Pa

cheers I hope this helped !!

3 0
2 years ago
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