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3 years ago

7

Suppose that you are standing on a train accelerating at 0.20g. What minimum coefficient of static friction must exist between y

our feet and the floor if you are not to slide.

1 answer:

umka21 [38]3 years ago

4 0

Friction force and train accelerating force are opposite in direction and equal in magnitude. Therefore m x a = u x m x g

As the mass is same, hence both are cancel out and remaining equation is

a = u x g

It’s given that a = 0.20g, so

0.20g = u x g

Cancel “g” on both sides;

0.20 = u

<span>Thus the coefficient of static friction = u = 0.20</span>

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Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.

jarptica [38.1K]

$y''-y'+y=\sin x$

The corresponding homogeneous ODE has characteristic equation $r^2-r+1=0$ with roots at $r=\dfrac{1\pm\sqrt3}2$ , thus admitting the characteristic solution

$y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x$

For the particular solution, assume one of the form

$y_p=a\sin x+b\cos x$

${y_p}'=a\cos x-b\sin x$

${y_p}''=-a\sin x-b\cos x$

Substituting into the ODE gives

$(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x$

$-b\cos x+a\sin x=\sin x$

$\implies a=1,b=0$

Then the general solution to this ODE is

$\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}$

$y''-3y'+2y=e^x\sin x$

$\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2$

$\implies y_c=C_1e^x+C_2e^{2x}$

Assume a solution of the form

$y_p=e^x(a\sin x+b\cos x)$

${y_p}'=e^x((a+b)\cos x+(a-b)\sin x)$

${y_p}''=2e^x(a\cos x-b\sin x)$

Substituting into the ODE gives

$2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x$

$-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x$

$\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12$

so the solution is

$\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}$

$y''+y=x\cos(2x)$

$r^2+1=0\implies r=\pm i$

$\implies y_c=C_1\cos x+C_2\sin x$

Assume a solution of the form

$y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)$

${y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)$

Substituting into the ODE gives

$(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)$

$-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)$

$\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49$

so the solution is

$\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}$

7 0

2 years ago

In the adjoining figure , APB and AQC are equilateral triangles. Prove that PC = BQ. ( Hint : <img src="https://tex.z-dn.net/?f=

just olya [345]

Answer:

See Below.

Step-by-step explanation:

Statements: Reasons:

$\displaystyle 1)\text{ } \Delta APB \text{ and } \Delta AQC \text{ are equilateral triangles}$ Given

$\displaystyle 2) \text{ } m \angle PAB = 60$ Definition of equilateral.

$3)\text{ } m \angle QAC = 60$ Definition of equilateral.

$4)\text{ } m\angle PAB = m\angle QAC$ Substitution

$5)\text{ } m\angle PAC=m\angle PAB+m\angle BAC$ Angle Addition

$\displaystyle 6)\text{ } m\angle QAB=m\angle QAC+m\angle BAC$ Angle Addition

$7)\text{ } m\angle QAB=m\angle PAB+m\angle BAC$ Substitution

$\displaystyle 8)\text{ } m\angle PAC=m\angle QAB$ Substitution

$9)\text{ } PA=BA$ Definition of equilateral

$10)\text{ } AC=AQ$ Definition of equilateral

$\displaystyle 11)\text{ } \Delta PAC \cong \Delta BAQ$ Side-Angle-Side Congruence*

$\displaystyle 12)\text{ } PC=BQ$ CPCTC

* SAS Congruence:

PA = BA

∠PAC = ∠QAB

AC = AQ

6 0

2 years ago

Read 2 more answers

What is the equation, in standard form, of a parabola that models the values in the table?

love history [14]

Answer:

<u> $Y=4x^2+3x-6$ </u>

Step-by-step explanation:

For the standard form equation to model the values in the table, each value of x in the table should give the matching the y value when substituted into the equation. We will test each equation:

<u> $Y=3x^2+4x-6$ for (-2,4)</u>

$Y=3(-2)^2+4(-2)-6=3(4)+-8-6=12+-8-6=-2\\$

This does not give 4 as the answer and is not a solution.

<u> $Y=4x^2+3x-6$ for (-2,4)</u>

$Y=4(-2)^2+3(-2)-6=4(4)+-6-6=16+-6-6=-4\\$

This does give 4 as the answer and is a possible solution.

<u> $Y=4x^2-3x-6$ for (-2,4)</u>

$Y=4(-2)^2-3(-2)-6=4(4)+6-6=16+6-6=16\\$

This does not give 4 as the answer and is not a solution.

<u> $Y=-4x^2-3x-6$ for (-2,4)</u>

$Y=-4(-2)^2-3(-2)-6=-4(4)+6-6=-16+6-6=-16\\$

This does not give 4 as the answer and is not a solution.

The only possible solution is <u> $Y=4x^2+3x-6$ </u>

3 0

3 years ago

Product of two integers is -6 and There sum is 1

dedylja [7]

Answer:

3 and -2

Step-by-step explanation:

xy = -6

x + y = 1

let 'x' = y-1

y(y-1) = -6

y² - y + 6 = 0

(y-3)(y+2) = 0

y = 3 and -2

7 0

3 years ago

Pls help I don’t understand this

Thepotemich [5.8K]

20% I think I’m sorry

5 0

2 years ago