Answer:
The rock's final speed at the required altitude will be 42.24 m/s.
Explanation:
Let's start by finding the initial vertical speed.
Vertical Speed = 1.61 * Sin (53.2°)
Vertical Speed = 0.8 m/s
We want to know the speed of the rock when it is at an altitude of 91 km.
The total displacement of the rock from its starting position will thus be equal to -91 km
We can use this in the following equation:
t = 4.3918 seconds
Thus it takes 4.3918 seconds to reach the required altitude. We can now find the speed as follows:
Thus the rock's final speed at the required altitude will be 42.24 m/s.
Answer:
v = 29.4m/s
Explanation:
Since the ball is dropped at rest,
u = 0m/s
a = 9.81m/s²
Using
v = u + at
After 3 seconds,
v = 0 + (9.81)(3)
v = 29.4m/s
D = 1/2 g t^2. It works out to 44.1 meters.
Answer:
ω = 630.2663 = 630[rad/s]
Explanation:
Solution:
- We can tackle this question by simple direct proportion relation between angular speed for the disk to rotate a cycle that constitutes 20 holes. We will use direct relation with number of holes per cycle to compute the revolution per seconds i.e frequency of speed f.
1rev(20 hole) -> 20(cycle)/rev
2006.2(cycle) -> f ?
f = 2006.2/20 = 100.31rev at second
- The relation between angular frequency and angular speed is given by:
ω = 2πf
ω = 2*3.14*100.31
ω = 630.2663 = 630[rad/s]
