Answer:
B.)Angular momentum is always conserved
Explanation:
Angular momentum is given by:
where
m is the mass of the object
v is its speed
r is the distance between the object and the centre of its circular trajectory
In absence of external torques, angular momentum is always conserved. That means that for the spinning star, if its radius r decreases (because it shrinks), in order for L (the angular momentum) to be conserved, the speed (v) must increases, therefore the spinning star speeds up.
So, the correct choice is
B.)Angular momentum is always conserved
This question involves the concepts of orbital velocity and orbital radius.
The orbital velocity of ISS must be "7660.25 m/s".
The orbital velocity of the ISS can be given by the following formula:
where,
v = orbital velocity = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of Earth = 5.97 x 10²⁴ kg
R = orbital radius = radius of earth + altitude = 63.78 x 10⁵ m + 4.08 x 10⁵ m
R = 67.86 x 10⁵ m
Therefore,
<u>v = 7660.25 m/s</u>
Learn more about orbital velocity here:
Answer:
<em>The final speed of the second package is twice as much as the final speed of the first package.</em>
Explanation:
<u>Free Fall Motion</u>
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:
And the distance traveled downwards is:
If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:
Replacing into the first equation:
Rationalizing:
Let's call v1 the final speed of the package dropped from a height H. Thus:
Let v2 be the final speed of the package dropped from a height 4H. Thus:
Taking out the square root of 4:
Dividing v2/v1 we can compare the final speeds:
Simplifying:
The final speed of the second package is twice as much as the final speed of the first package.