Answer:
0.231 m/s
Explanation:
m = mass attached to the spring = 0.405 kg
k = spring constant of spring = 26.3 N/m
x₀ = initial position = 3.31 cm = 0.0331 m
x = final position = (0.5) x₀ = (0.5) (0.0331) = 0.01655 m
v₀ = initial speed = 0 m/s
v = final speed = ?
Using conservation of energy
Initial kinetic energy + initial spring energy = Final kinetic energy + final spring energy
(0.5) m v₀² + (0.5) k x₀² = (0.5) m v² + (0.5) k x²
m v₀² + k x₀² = m v² + k x²
(0.405) (0)² + (26.3) (0.0331)² = (0.405) v² + (26.3) (0.01655)²
v = 0.231 m/s
A motion security line is a system that is used to detect motion.
The input for the system is MOTION and the output is LIGHT. That is, when the system detects motion it switch on light.
Remember, an Input is the information that was inserted into a system while the output is the result of the processed information.
Answer:
The coefficient of kinetic friction between the puck and the ice is 0.11
Explanation:
Given;
initial speed, u = 9.3 m/s
sliding distance, S = 42 m
From equation of motion we determine the acceleration;
v² = u² + 2as
0 = (9.3)² + (2x42)a
- 84a = 86.49
a = -86.49/84
|a| = 1.0296
= ma
where;
Fk is the frictional force
μk is the coefficient of kinetic friction
N is the normal reaction = mg
μkmg = ma
μkg = a
μk = a/g
where;
g is the gravitational constant = 9.8 m/s²
μk = a/g
μk = 1.0296/9.8
μk = 0.11
Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11
Answer:
Take-off velocity = v = 81.39[m/s]
Explanation:
We can calculate the takeoff speed easily, using the following kinematic equation.

where:
a = acceleration = 4[m/s^2]
x = distance = 750[m]
vi = initial velocity = 25 [m/s]
vf = final velocity
![v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D%5Csqrt%7B%2825%29%5E%7B2%7D%2B%282%2A4%2A750%29%20%7D%20%5C%5Cv_%7Bf%7D%3D81.39%5Bm%2Fs%5D)
The statement to every reacting there is, there is a opposite and same reacting.
hope it helps