The distance formula:d = v i · t + a · t² / 2Sir G.: ( v i = 0 )
d G = 0 + 0.3 · t² / 2Sir A. : d A = 0 + 0.2 t² / 288 m = 0.3 t² / 2 + 0.2 t² / 2 / · 2 ( multiple both sides by 2 )176 = 0.3 t² + 0.2 t²176 = 0.5 t²t² = 176 : 0.5t² = 352t = √352t = 18.76 sd G. = 0.3 · 18.76² / 2 = 0.3 · 352 / 2 = 52.8 mAnswer: The knights collide at 52.8 m relative to Sir George`s starting point.
Particles similar to a water wave will travel away from the source. On the Earth's surface this would be shown as an aftershock. On the surface of water, the waves would travel in ripples.
Hello =D
This problem is about cinematic
So
V = 45 mi/h
t = 2 h
Then
V= X/t
X = V*t
Then
X = (45)*(2)
X = 90 mi
Best regards
So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.
<h3>Introduction</h3>
Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :
With the following condition :
= angular frequency (rad/s)
= change of angle value (rad)- t = interval of the time (s)
<h3>Problem Solving</h3>
We know that :
- = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
- t = interval of the time = 54.9 s.
What was asked :
- = angular frequency = ... rad/s
Step by step :
<h3>Conclusion :</h3>
So, the angular frequency of the blades approximately 36.43π rad/s.
Area near a sea having flat land and low relief
