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3 years ago

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When preparing for a rocket launch, the mission control center uses the phrase "T minus" before liftoff. ...T minus 3, T minus 2

, T minus 1, ... After the rocket has launched, "T plus" is used while the rocket is in flight. ...T plus 1, T plus 2, T plus 3, ... When does the rocket launch? What does "T" represent?

1 answer:

yKpoI14uk [10]3 years ago

6 0

Answer: The "T" in rocket launch

Explanation:

This tells the Mission Control What will be happening

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Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

AURORKA [14]

Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

$r=\sqrt{61}$

$\theta=tan^{-1}(\frac{6}{-5})$

$\theta=tan^{-1}(-1.2)$

$\theta=-50.2$ °

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

$\theta=360-50.2$

$\theta=$ 309.8°

Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

For point C:

$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

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You start pushing a suitcase full of clothes in the horizontal direction with a force of 25.0 newtons. The weight of the suitcas

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Answer:

Distance traveled will be 5.6307 m

Explanation:

Time t = 3 sec

We have given force F = 25 N

We know that force is given by F = ma

So ma = 25 -----------eqn 1

Weight is given by W = 196 N

We know that weight is given by W = mg

So mg = 196 -----------------eqn 2

From equation 1 and equation 2 $\frac{a}{g}=\frac{25}{196}$

$a=1.2512m/sec^2$

Initial velocity is given as 0 so u = 0 m/sec

From second equation of motion $s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 1.2512\times 3^2=5.6307m$

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For an object in uniform circular motion, what can you say about the directions of the velocity, acceleration, and net force vec

zavuch27 [327]

Answer: b) The velocity vector is perpendicular to the acceleration vector; the acceleration vector is parallel to the net force vector.

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Lukalu is rappelling off a cliff. The parametric equations that describe her horizontal and vertical position as a function of t

andre [41]

Answer:

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y(t) = 0

So we find:

$0=-16t^2 +100\\16t^2 = 100\\t=\sqrt{\frac{100}{16}}=2.5 s$

The horizontal distance of Lukalu instead will be given by the equation for the horizontal position, substituting t = 2.5 s:

$x=8t = 8 \cdot 2.5 s =5 m$

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Franklin’s hometown is flooding due to a severe storm. He is evacuating the area in his vehicle. What should he do if he comes a

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Franklin must not drive through a flood for there may no road at all under the water, unless he is familiar with the road and the flowing water is below one foot.

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