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.
Answer: 3 radians/meter.
Explanation:
The general sinusoidal function will be something like:
y = A*sin(k*x - ω*t) + C
Where:
A is the amplitude.
k is the wave number.
x is the spatial variable
ω is the angular frequency
t is the time variable.
C is the mid-value.
The rule that we can use to solve this problem, is that the argument of the sin( ) function must be in radians (or in degrees)
Then if x is in meters, the wave-number must be in radians/meters, so when these numbers multiply the "meters" part is canceled.
Then for the case of the function:
y(x,t) = 0.1 sin(3x + 10t)
Where x is in meters, the units of the wave number (the 3) must be in radians/meters. Then the angular wave number is 3 radians/meter.
Answer:
K_{total} = 19.4 J
Explanation:
The total kinetic energy that is formed by the linear part and the rotational part is requested
let's look for each energy
linear
= ½ m v²
rotation
= ½ I w²
the moment of inertia of a solid sphere is
I = 2/5 m r²
we substitute
= ½ mv² + ½ I w²
angular and linear velocity are related
v = w r
we substitute
K_{total} = ½ m w² r² + ½ (2/5 m r²) w²
K_{total} = m w² r² (½ + 1/5)
K_{total} = 
m w² r²
let's calculate
K_{total} = 6.40 16.0² 0.130²
K_{total} = 19.4 J
A. The inner core heat produce stuff
