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3 years ago

What is not an example of kinetic energy?

Mathematics

2 answers:

victus00 [196]3 years ago

6 0

Answer:

this is ur answer:

Step-by-step explanation:

Moving Car. Moving cars possess some amount of energy !

hope it helps !

Anestetic [448]3 years ago

5 0

Answer:

Moving Car. Moving cars possess some amount of kinetic energy. ...

Bullet From a Gun. A bullet fired from a gun has very high kinetic energy, and, so, it can easily penetrate any object. ...

Flying Airplane. ...

Walking & Running. ...

Cycling. ...

Rollercoasters. ...

Cricket Ball.

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Which coordinate plane contains the points (2, 2.5) and ( -4, -4)

ivann1987 [24]

Answer:

Step-by-step explanation:

That is the only answer provided.

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3 years ago

A football field is rectangular and measures 360 feet by 160 feet. A flying disk is randomly tossed onto the field. If it is equ

kaheart [24]

The dimensions if the field are: the length is 360 feet; the width is 160 feet.

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3 years ago

Draw a model to represent the following problem 42 divide 7 Answer

hodyreva [135]

We have to draw the model for 42 divide 7 . In order to draw the model we use below steps.

Draw 7 boxes to represent 7 groups.

Assume 42 as the number which represents the total of these 7 groups.

Put a question mark at the first box.

Attached is the model to represents 42 divide 7 .

In the question mark we should put 6. Hence, when we add 6 to 7 times that will be equal to 42.

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3 years ago

Use the variation of parameters method to solve the DR y" + y' - 2y = 1

postnew [5]

Answer:

$y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Step-by-step explanation:

As given in question, we have to find the solution of differential equation

$y"+y'-2y=1$

by using the variation in parameter method.

From the above equation, the characteristics equation can be given by

$D^2+D-2\ =\ 0$

$=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}$

$=>\ D=\ -2\ or\ 1$

Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by

$y_c(t)\ =\ C_1e^{-2t}+C_2e^t$

Let's assume that

$y_1(t)=e^{-2t}$ $y_2(t)=e^t$

$=>\ y'_1(t)=-2e^{-2t}$ $y'_2(t)=e^t$

and g(t)=1

Now, the Wronskian can be given by

$W=y_1(t).y'_2(t)-y'_1(t).y_2(t)$

$=e^{-2t}.e^t-e^t(-e^{-2t})$

$=e^{-t}+2e^{-t}$

$=3e^{-t}$

Now, the particular solution can be given by

$y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}$

$=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}$

$=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}$

$=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}$

$=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Now, the complete solution of the given differential equation can be given by

$y(t)\ =\ y_c(t)+y_p(t)$

$=C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

5 0

3 years ago

Does the following table show a proportional relationship between the variables X and y?

Lana71 [14]

Answer:

Step-by-step explanation:

5 0

3 years ago