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.
CuCl2 will have dipole-dipole forces the strongest attraction that will be present. For CuSO4, io- dipole forces would be the strongest and for HF, hydrogen bonding would be the strongest force present. Dipole-dipole are permanent dipoles where two polar molecules interact. For ion-dipole, ions are the contributing factor of such and for hydrogen bonding, it is the bonding of hydrogen atom with a halogen.
Given the time and the horizontal velocity, we can simply
compute for the distance how far the ball travelled using the formula:
distance = velocity * time
<span>Since velocity is in units of m/s and time is seconds,
therefore we can directly get a unit in meters.</span>
Answer:
Part a)
Part b)
Part c)
Explanation:
Part a)
as we know that angular acceleration of the wheel is given as
now the radius of the wheel is given as
R = 3.21 cm
so the tangential acceleration is given as
Part b)
frequency of the wheel at maximum speed is given as
now we know that
now radial acceleration is given as
Part c)
total angular displacement of the point on rim is given as
here we know that
now angular displacement will be
now the distance moved by the point on the rim is given as