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RoseWind [281]
3 years ago
11

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?
Physics
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°)

3 0
3 years ago
You start pushing a suitcase full of clothes in the horizontal direction with a force of 25.0 newtons. The weight of the suitcas
sasho [114]

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

7 0
3 years ago
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.

Explanation: A change in velocity creates an acceleration. As the object rotates through the circular path it is constantly changing direction, and hence accelerating, which causes a constant force to act upon the object. This Force acts towards the center of curvature, directly toward the axis of rotation in a direction parallel to the acceleration of the body along the path. Because the object is moving perpendicular to the force, the path followed by the object is a circular one. Hence the velocity of the object is perpendicular to the acceleration.

3 0
3 years ago
Lukalu is rappelling off a cliff. The parametric equations that describe her horizontal and vertical position as a function of t
andre [41]

Answer:

2.5 s, 5 m

Explanation:

The equations for the horizontal and vertical position of Lukalu are:

x(t) = 8t\\y(t) = -16t^2 + 100

we can find the time it takes her to reach the ground by requiring that the vertical position becomes zero:

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

4 0
3 years ago
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
Mrac [35]

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. 

Moreover, if negotiating a flooded section of road, he must drive in the middle where the water will be at its shallowest and he must not drive through water against approaching vehicle to consider other drivers.

4 0
3 years ago
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