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.
28.1 grams is 0.0281 kg.
43.278 kg - 0.0281 kg = 43.2499 or if you round -> 43.250
V = 331 m/s
Wavelength = velocity of sound/ frequency
Frequency = velocity of sound / wavelength = 331 m/s : 0.6 m
Frequency = 551.67 s^-1 = 551.67 Hz
Answer:
50J
Explanation:
Potential energy = mass (kg) * gravitational accelerations (10N on earth) * height
250 g in kg = 0.25
0.25 * 10 * 20 = 50 J