Barometer is taken from the base to the top of a 279-m tower. assuming the density of air is 1.29 kg/m3, what is the measured ch
1 answer:
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For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:
1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J
2) The potential energy at <u>point A</u> is 19620 J
3) The kinetic energy at <u>point B</u> is 10010 J
4) The potential energy at <u>point C</u> is zero
5) The kinetic energy at <u>point C</u> is 19820 J
6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s
1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:
Where:
KE: is the kinetic energy = (1/2)mv₀²
m: is the mass of the roller coaster = 100 kg
v₀: is the initial velocity = 2 m/s
PE: is the potential energy = mgh
g: is the acceleration due to gravity = 9.81 m/s²
h: is the height = 20 m
The<em> total energy</em> is:
Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.
2) The <em>potential energy</em> at point A is:
Then, the potential energy at <u>point A</u> is 19620 J.
3) The <em>kinetic energy</em> at point B is the following:
Since
we have:
Hence, the kinetic energy at <u>point B</u> is 10010 J.
4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.
5) The <em>kinetic energy</em> of the roller coaster at point C is:
Therefore, the kinetic energy at <u>point C</u> is 19820 J.
6) The <em>velocity</em> of the roller coaster at point C is given by:
Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.
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Answer:
At the closest point
Explanation:
We can simply answer this question by applying Kepler's 2nd law of planetary motion.
It states that:
"A line connecting the center of the Sun to any other object orbiting around it (e.g. a comet) sweeps out equal areas in equal time intervals"
In this problem, we have a comet orbiting around the Sun:
- Its closest distance from the Sun is 0.6 AU
- Its farthest distance from the Sun is 35 AU
In order for Kepler's 2nd law to be valid, the line connecting the center of the Sun to the comet must move slower when the comet is farther away (because the area swept out is proportional to the product of the distance and of the velocity: , therefore if r is larger, then v (velocity) must be lower).
On the other hand, when the the comet is closer to the Sun the line must move faster (, if r is smaller, v must be higher). Therefore, the comet's orbital velocity will be the largest at the closest distance to the Sun, 0.6 A.