Kinetic energy = 1/2 m v^2 = 1/2 x1.5 x10^-3 x 0.36
The solution for this problem is:
r = [(2.90 + 0.0900t²) i - 0.0150t³ j] m/s²
this is for t in seconds and r in meters
v = dr/dt = [0.180t i - 0.0450t² j] m/s²
tan(-36.0º) = -0.0450t² / 0.180t
0.7265 = 0.25t
t = 2.91 s is the velocity vector of the insect
Answer:
189 m/s
Explanation:
The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.
So, F = W
mv²/r = mg
v² = gr
v = √gr where v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m
So, v = √gr
v = √(9.8 m/s² × 3.63 × 10³ m)
v = √(35.574 × 10³ m²/s²)
v = √(3.5574 × 10⁴ m²/s²)
v = 1.89 × 10² m/s
v = 189 m/s
The distance of the canoeist from the dock is equal to length of the canoe, L.
<h3>
Conservation of linear momentum</h3>
The principle of conservation of linear momentum states that the total momentum of an isolated system is always conserved.
v(m₁ + m₂) = m₁v₁ + m₂v₂
where;
v is the velocity of the canoeist and the canoe when they are together
- u₁ is the velocity of the canoe
- u₂ velocity of the canoeist
- m₁ mass of the canoe
- m₂ mass of the canoeist
<h3>Distance traveled by the canoeist</h3>
The distance traveled by the canoeist from the back of the canoe to the front of the canoe is equal to the length of the canoe.
Thus, the distance of the canoeist from the dock is equal to length of the canoe, L.
Learn more about conservation of linear momentum here: brainly.com/question/7538238
