Answer:
When the parachute opens, the air resistance increases. The skydiver slows down until a new, lower terminal velocity is reached.
The orbital period could be obtained from the question as 8.98 * 10^7 m.
<h3>What is the orbital period?</h3>
We know that the solar system is composed of the sun and the planets. The planets are known to move around the sun in concentric circles. Following the heliocentric model of the solar system, the sun is at the center of the solar system at all times and all the other planets tend to move round the sun at a good distance that is appropriate for each.
Now we have that;
T = √4π^2r^3/GM
T = period of the orbit
r = radius of the orbit
G = gravitational constant
M = mass of the planet
Hence, we have;
R = ∛T^2GM/4π^2
R =∛ (26 * 60 * 60)^2 * 6.67 * 10^-11 * 4.9 * 10^25/ 4 * (3.142)^2
R = ∛2.86 * 10^25/39.49
R = 8.98 * 10^7 m
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Given:
altitude, x = 1 mile
speed, v = 560 mi/h
distance from the station, x = 4 mi
Solution:
To find the rate,
Now, from the right angle triangle in fig 1.
Applying pythagoras theorem:
differentiating the above eqn w.r.t 't' :
(1)
Now, putting values in eqn (1):
The rate at which distance from plane to station is increasing is:
<span>C: 8 cm
Assuming no friction, since the ball is starting with an elevation of 20 cm, it will end up with an elevation of 20 cm. So let's just calculate the result.
"slope of 5 cm of rise for each 2 cm of run". So for every 5 cm the ball rises, it will go 2 cm horizontally. Since we know the ball will have a total rise of 20 cm, that gives us
20 cm / 5 cm * 2 cm = 8 cm
And looking at the available answers, 8 cm is sitting right there. So the answer is
C: 8 cm</span>