<u>ALL of the following work assumes NO AIR RESISTANCE:</u>
1). an object moving under the influence of only gravity, and not in orbit; its horizontal velocity is constant, and its vertical motion is accelerated downward at 9.8 m/s²
2). a parabola
3). Horizontal: velocity is constant, acceleration is zero. . . . Vertical: acceleration is 9.8 m/s² downward, velocity depends on whether it was launched, thrown up, thrown down, dropped, etc.
4). a). the one that was thrown horizontally; b). both hit the ground at the same time; c). both hit the ground with the same vertical velocity
5). a). zero; b). zero; c). gravity ... 9.8 m/s² down; d). 3.06 seconds; e). 4.38 m/s; f). 30 m/s g). no; gravity has no effect on horizontal motion
6). a). 1.8 seconds; b). 13.1 meters; c). 17.6 m/s down; d). 7.3 m/s; gravity has no effect on horizontal motion
7). 45 m/s
8). without air resistance, the ball is traveling horizontally at 13 km/hr, and it lands back in your hand
9). a). 4.49 m/s; b). 29.7 m/s
10). 7.24 meters
11). 700 meters
12). A). 103.7 meters ( ! she's in big trouble ! ); B). 17.5 meters
Answer:

t'=1.1897 μs
Explanation:
First we will calculate the velocity of micrometeorite relative to spaceship.
Formula:

where:
v is the velocity of spaceship relative to certain frame of reference = -0.82c (Negative sign is due to antiparallel track).
u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)
u' is the relative velocity of micrometeorite with respect to spaceship.
In order to find u' , we can rewrite the above expression as:


u'=0.9806c
Time for micrometeorite to pass spaceship can be calculated as:

(c = 3*10^8 m/s)


t'=1.1897 μs
Answer: D. wealthy
Explanation: on the e2020 test its right
We know that
g = LcosΘ
<span>where g, L and Θ are centripetal gravity length, and angle of object
</span><span>ω² = g/LcosΘ </span>
<span>ω = √(g / LcosΘ) </span>
Explanation:
We know that the sky appears to us like a sphere called as celestial sphere which appears to rotate around an imaginary axis because of Earth's rotation. Since the axis cuts the celestial sphere at celestial poles all the object seems to circle around the celestial poles.
Condition 1: The stars rise and set perpendicular to the horizon
The observer is at the equator
Condition 2: The stars circle the sky parallel to the horizon
The observer is at the Pole of the Earth
Condition 3: The celestial equator passes through the zenith
The observer is at the equator
Condition 4: In the course of a year, all stars are visible
The observer is at the equator
Condition 5: The Sun rises on March 21 and does not set until September 21 (ideally)
The observer is at North Pole