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3 years ago

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A box is tossed upward from a building that is 18 meters high, at a speed of 9 meters per second. What is the height of the box

at 1 second if the height is given in meters by h(t)=−4.9t^2+9t+18 where t is seconds?

2 answers:

Anon25 [30]3 years ago

7 0

Answer:

22.1 meters

Step-by-step explanation:

The equation for the height of the box is given by:

$h(t)=-4.9t^2+9t+18$ where t is time in seconds.

After 1 second,

The height of the box

$h(1)=-4.9(1)^2+9(1)+18\\=-4.9+9+18\\=22.1 \:meters$

Strike441 [17]3 years ago

3 0

Answer:

22.1 meters

Step-by-step explanation:

To calculate the height of the box at 1 second after it was tossed upward, we just need to use the equation of the height that the question gave us - h(t) - using the time t = 1:

h(t) = -4.9*t^2 + 9*t + 18

h(1) = -4.9*1^2 + 9*1 + 18

h(1) = -4.9 + 9 + 18 = 22.1 meters

The height of the box at 1 second is 22.1 meters.

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3 years ago

Let C be the boundary of the region in the first quadrant bounded by the x-axis, a quarter-circle with radius 9, and the y-axis,

rewona [7]

Solution :

Along the edge $C_1$

The parametric equation for $C_1$ is given :

$x_1(t) = 9t , y_2(t) = 0 \ \ for \ \ 0 \leq t \leq 1$

Along edge $C_2$

The curve here is a quarter circle with the radius 9. Therefore, the parametric equation with the domain $0 \leq t \leq 1$ is then given by :

$x_2(t) = 9 \cos \left(\frac{\pi }{2}t\right)$

$y_2(t) = 9 \sin \left(\frac{\pi }{2}t\right)$

Along edge $C_3$

The parametric equation for $C_3$ is :

$x_1(t) = 0, \ \ \ y_2(t) = 9t \ \ \ for \ 0 \leq t \leq 1$

Now,

x = 9t, ⇒ dx = 9 dt

y = 0, ⇒ dy = 0

$\int_{C_{1}}y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

And

$x(t) = 9 \cos \left(\frac{\pi}{2}t\right) \Rightarrow dx = -\frac{7 \pi}{2} \sin \left(\frac{\pi}{2}t\right)$

$y(t) = 9 \sin \left(\frac{\pi}{2}t\right) \Rightarrow dy = -\frac{7 \pi}{2} \cos \left(\frac{\pi}{2}t\right)$

Then :

$\int_{C_1} y^2 x dx + x^2 y dy$

$=\int_0^1 \left[\left( 9 \sin \frac{\pi}{2}t\right)^2\left(9 \cos \frac{\pi}{2}t\right)\left(-\frac{7 \pi}{2} \sin \frac{\pi}{2}t dt\right) + \left( 9 \cos \frac{\pi}{2}t\right)^2\left(9 \sin \frac{\pi}{2}t\right)\left(\frac{7 \pi}{2} \cos \frac{\pi}{2}t dt\right) \right]$

$=\left[-9^4\ \frac{\cos^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} -9^4\ \frac{\sin^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} \right]_0^1$

= 0

And

x = 0, ⇒ dx = 0

y = 9 t, ⇒ dy = 9 dt

$\int_{C_3} y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

Therefore,

$\oint y^2xdx +x^2ydy = \int_{C_1} y^2 x dx + x^2 x dx+ \int_{C_2} y^2 x dx + x^2 x dx+ \int_{C_3} y^2 x dx + x^2 x dx$

= 0 + 0 + 0

Applying the Green's theorem

$x^2 +y^2 = 81 \Rightarrow x \pm \sqrt{81-y^2}$

$\int_C P dx + Q dy = \int \int_R\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx dy$

Here,

$P(x,y) = y^2x \Rightarrow \frac{\partial P}{\partial y} = 2xy$

$Q(x,y) = x^2y \Rightarrow \frac{\partial Q}{\partial x} = 2xy$

$\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) = 2xy - 2xy = 0$

Therefore,

$\oint_Cy^2xdx+x^2ydy = \int_0^9 \int_0^{\sqrt{81-y^2}}0 \ dx dy$

$= \int_0^9 0\ dy = 0$

The vector field F is = $y^2 x \hat i+x^2 y \hat j$ is conservative.

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3 years ago

Find the volume of the solid whose base is the region in the first quadrant bounded by y=x^3, y=1, and the y axis whose cross se

vodomira [7]

The region in question is the set

$R = \left\{ (x, y) : 0 \le x \le 1 \text{ and } x^3 \le y \le 1 \right\}$

or equivalently,

$R = \left\{ (x, y) : 0 \le y \le 1 \text{ and } 0 \le x \le \sqrt[3]{y} \right\}$

Cross sections are taken perpendicular to the y-axis, which means each section has a base length equal to the horizontal distance between the curve y = x³ and the line x = 0 (the y-axis). This horizontal distance is given by

y = x³ ⇒ x = ∛y

so that each triangular cross section has side length ∛y.

The area of an equilateral triangle with side length s is √3/4 s², so each cross section contributes an infinitesimal area of √3/4 ∛(y²).

Then the volume of this solid is

$\displaystyle \frac{\sqrt3}4 \int_0^1 \sqrt[3]{y^2} \, dy = \frac{\sqrt3}4 \int_0^1 y^{2/3} \, dy = \frac{\sqrt3}4\cdot\frac35 y^{5/3} \bigg|_0^1 = \boxed{\frac{3\sqrt3}{20}}$

I've attached some sketches of the solid with 16 and 64 such cross sections to give an idea of what this solid looks like.

3 0

2 years ago

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marysya [2.9K]

Answer:

12 20 22

Step-by-step explanation:

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3 years ago