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3 years ago

Please help me fast I don’t have much time to answer so I need help!!!!

Mathematics

1 answer:

Sonja [21]3 years ago

4 0

Answer:

Step-by-step explanation:

you click the first one then you click the second one then you click the fourth one

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A small rocket is fired from a launch pad 10 m above the ground with an initial velocity left angle 250 comma 450 comma 500 righ

jonny [76]

Let $\vec r(t),\vec v(t),\vec a(t)$ denote the rocket's position, velocity, and acceleration vectors at time $t$ .

We're given its initial position

$\vec r(0)=\langle0,0,10\rangle\,\mathrm m$

and velocity

$\vec v(0)=\langle250,450,500\rangle\dfrac{\rm m}{\rm s}$

Immediately after launch, the rocket is subject to gravity, so its acceleration is

$\vec a(t)=\langle0,2.5,-g\rangle\dfrac{\rm m}{\mathrm s^2}$

where $g=9.8\frac{\rm m}{\mathrm s^2}$ .

a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,

$\vec v(t)=\left(\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du\right)\dfrac{\rm m}{\rm s}$

$\vec v(t)=\left(\langle250,450,500\rangle+\langle0,2.5u,-gu\rangle\bigg|_0^t\right)\dfrac{\rm m}{\rm s}$

(the integral of 0 is a constant, but it ultimately doesn't matter in this case)

$\boxed{\vec v(t)=\langle250,450+2.5t,500-gt\rangle\dfrac{\rm m}{\rm s}}$

and

$\vec r(t)=\left(\vec r(0)+\displaystyle\int_0^t\vec v(u)\,\mathrm du\right)\,\rm m$

$\vec r(t)=\left(\langle0,0,10\rangle+\left\langle250u,450u+1.25u^2,500u-\dfrac g2u^2\right\rangle\bigg|_0^t\right)\,\rm m$

$\boxed{\vec r(t)=\left\langle250t,450t+1.25t^2,10+500t-\dfrac g2t^2\right\rangle\,\rm m}$

b. The rocket stays in the air for as long as it takes until $z=0$ , where $z$ is the $z$ -component of the position vector.

$10+500t-\dfrac g2t^2=0\implies t\approx102\,\rm s$

The range of the rocket is the distance between the rocket's final position and the origin (0, 0, 0):

$\boxed{\|\vec r(102\,\mathrm s)\|\approx64,233\,\rm m}$

c. The rocket reaches its maximum height when its vertical velocity (the $z$ -component) is 0, at which point we have

$-\left(500\dfrac{\rm m}{\rm s}\right)^2=-2g(z_{\rm max}-10\,\mathrm m)$

$\implies\boxed{z_{\rm max}=125,010\,\rm m}$

7 0

3 years ago

-5 / 6 - 1 / x equals negative two thirds

anyanavicka [17]

$x = - 6$

I think it's help

6 0

4 years ago

I JUST NEED #13 please and thnx

emmainna [20.7K]

$\bf \cfrac{18}{32}\implies \cfrac{2\cdot 3\cdot 3}{2\cdot 2\cdot 2\cdot 2\cdot 2}\implies \cfrac{3\cdot 3}{2\cdot 2\cdot 2\cdot 2}\implies \boxed{\cfrac{9}{16}}\qquad \checkmark \\\\[-0.35em] ~\dotfill\\\\ \cfrac{27}{48}\implies \cfrac{3\cdot 3\cdot 3}{2\cdot 2\cdot 2\cdot 2\cdot 3}\implies \cfrac{3\cdot 3}{2\cdot 2\cdot 2\cdot 2}\implies \boxed{\cfrac{9}{16}}\qquad \checkmark$

3 0

3 years ago

On my brother's 14th birthday, he was 160 cm tall. Within exactly one year, his height increased by 6%. How tall was he on his 1

Charra [1.4K]

Answer:

170 cm

Step-by-step explanation:

160+.06*160=160+9.6=169.6

8 0

3 years ago

What is the transformation for y = lxl + 3 ?

anyanavicka [17]

Answer: Center: (0,0)

Vertices: ( √3 , 0 ) , ( − √3 , 0 )

Foci: ( √6 , 0 ) , ( − √6 , 0 )

Eccentricity: √2

Focal Parameter: √6/2

Asymptotes: y = x , y = − x

How to: Rewrite in vertex form and use this form to find the vertex ( h , k ) .

Have a great day and stay safe !

3 0

3 years ago