Answer:
Kepler's first law means that planets move around the Sun in elliptical orbits. An ellipse is a shape that resembles a flattened circle. ... It is zero for a perfect circle.
I can think of two possible and logical questions for the problem given. First, you can calculate for the maximum height reached by the blue ball. Second, you can compute the length of time for the two balls to be at the same height. If so, the solution are as follows:
When the object is thrown upwards or when the object is dropped from a height, the only force acting upon it is the gravitational force. Because of this, it simplifies equations of motion.
1. For the maximum height, the equation is
H = v₀²/2g
where
v₀ is the initial speed
g is the acceleration due to gravity equal to 9.81 m/s²
For the blue ball, v₀ = 21.8 m/s. Substituting the values:
H = (21.8 m/s)²/2(9.81m/s²)
H = 24.22 m
The maximum height reached by the blue ball is 24.22 m + 0.9 = 25.12 m.
2. For this, you equate the y values of both balls:
y for red ball = y for blue ball
v₀t + 0.5gt² = v₀t + 0.5gt²
(10.4 m/s)t + 0.5(9.81 m/s²)(t²) + 26.6 m = (21.8 m/s)t + 0.5(9.81 m/s²)(t²) + 0.9 m
Solving for t,
t = 2.25 seconds
Thus, the two balls would be at the same height after 2.25 seconds.
Yes, it's true.
But 2nd Newton Law always come to play when the horse is to move forward because obviously the forces interact antagonistically and mass has to be accounted for.
That's what I think. Hope it's right, all the best.
Answer:
6360 km
Explanation:
Use the kinematics equation 
. We are given t = 7.95 hours and a = 0 m/s^2 (constant speed means there is no acceleration). Solve for x.
<h2>
a) Displacement of penny = 1300 i + 2400 j - 640 k</h2><h2>b) Magnitude of his displacement = 2729.47 m</h2>
Explanation:
a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.
1300 m east = 1300 i
2400 m north = 2400 j
Drops the penny from a cliff 640 m high = -640 k
Displacement of penny = 1300 i + 2400 j - 640 k
b) Displacement of man for return trip = -1300 i - 2400 j
Magnitude of his displacement = 2729.47 m
