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natta225 [31]
3 years ago
12

The inclination (tilt) of an amusement park ride is accelerating at a rate of 2160\,\dfrac{\text{degrees}}{\text{min}^2}2160 min

2 degrees ​ 2160, start fraction, start text, d, e, g, r, e, e, s, end text, divided by, start text, m, i, n, end text, squared, end fraction. What is the ride's acceleration rate in \dfrac{\text{degrees}}{\text{s}^2} s 2 degrees ​ start fraction, start text, d, e, g, r, e, e, s, end text, divided by, start text, s, end text, squared, end fraction?
Mathematics
1 answer:
Amanda [17]3 years ago
7 0

Answer:

0.6 deg/s²

Step-by-step explanation:

Acceleration is the time rate of change of velocity, it is the ratio of velocity to time. The formula for acceleration is given by:

Acceleration = Change in velocity / time taken = (Final velocity - Initial velocity) / time

Given that the acceleration is 2160 deg/min², we have to convert it to deg/s²

1 min = 60 seconds

1 min² = 60² s² = 3600 s²

2160\ deg/min^2=\frac{2160\ deg}{1\ min^2*3600\ \frac{s^2}{min^2} }=0.6\ deg/s^2\\ \\2160\ deg/min^2=0.6\ deg/s^2

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Step-by-step explanation:

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leonid [27]

Answer:

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

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Step-by-step explanation:

\frac{dy}{dx}y=x^3+2x-5x+3

\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}

  • \mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)

1. \mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:

y'\:y=x^3+2x-5x+3

2. \mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}

yy'\:=x^3-3x+3

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5. \mathrm{Simplify}

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

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