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zhenek [66]
3 years ago
5

Consider the differential equation

Physics
1 answer:
Zielflug [23.3K]3 years ago
3 0
Yp(t) = A1 t^2 + A0 t + B0 t e(4t)

=> y ' = 2A1t + A0 + B0 [e^(4t) +4 te^(4t) ]

     y ' = 2A1t + A0 + B0e^(4t) + 4B0 te^(4t)

=> y '' = 2A1 + 4B0e(4t) + 4B0 [ e^(4t) + 4te^(4t)

     y '' = 2A1 + 4B0e^(4t) + 4B0e^(4t) + 16B0te^(4t)

Now substitute the values of y ' and y '' in the differential equation:

 
<span>y′′+αy′+βy=t+e^(4t)


</span> 2A1 + 4B0e^(4t) + 4B0e^(4t) + 16B0te^(4t) + α{2A1t + A0 + B0e^(4t) + 4B0 te^(4t) } + β{A1 t^2 + A0 t + B0 t e(4t)} = t + e^(4t)

Next, we equate coefficients

1) Constant terms of the left side = constant terms of the right side:

2A1+ 2αA0 = 0 ..... eq (1)

2) Coefficients of e^(4t) on both sides

8B0 + αB0 = 1 => B0 (8 + α) = 1 .... eq (2)

3) Coefficients on t

2αA1 + βA0 = 1 .... eq (3)


4) Coefficients on t^2

βA1 = 0 ....eq (4)

given that A1 ≠ 0 => β =0

5) terms on te^(4t)

16B0 + 4αB0 + βB0 = 0 => B0 (16 + 4α + β) = 0 ... eq (5)

Given that B0 ≠ 0 => 16 + 4α + β = 0

Use the value of β = 0 found previously

16 + 4α = 0 => α = - 16 / 4 = - 4.

Answer: α = - 4 and β = 0
  




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Consider the following distribution of objects: a 2.00-kg object with its center of gravity at (0, 0) m, a 2.20-kg object at (0,
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Answer:

body position 4 is (-1,133, -1.83)

Explanation:

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               y_cm = 1 /M   ∑ y_{i} mi

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give us the mass and position of this masses

body 1

m1 = 2.00 ka

x1 = 0 me

y1 = 0 me

body 2

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x2 = 0m

y2 = 5 m

body 3

m3 = 3.4 kg

x3 = 2.00 m

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body 4

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    x4=?

   y4=?

mass center position

x_cm = 0

y_cm = 0

let's apply to the equations of the initial part

X axis

    M = 2.00 + 2.20 + 3.40

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    0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)

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Axis y

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What is the change in thermal energy E if the coefficent of kinetic friction between the box and floor is .4 , the distance the
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This question can be solved using the concept of friction energy.

The thermal energy change is b "258.4 J".

The change in thermal energy will be equal to the friction energy produced during the motion of the box.

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where,

μ = coefficient of kinetic friction = 0.4

f = force applied = 38 N

d = distance traveled by the box = 17 m

Therefore,

E = (0.4)(38\ N)(17\ m)

<u>E = 258.4 J</u>

Learn more about friction energy here:

brainly.com/question/1343045?referrer=searchResults

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