Answer:
<em>k = 25.18 N/m</em>
Explanation:
<u>Simple Harmonic Oscillator</u>
It consists of a weight attached to one end of a spring being allowed to move forth and back.
If m is the mass of the weight and k is the constant of the spring, the period of the oscillation is given by:
If the period is known, we can find the value of the constant by solving for k:
Substituting the given values m=5 Kg and T=2.8 seconds:
k = 25.18 N/m
Q: ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself off in the opposite direction at 60 m/s. What is the magnitude of the average force the wall exerts on Ken during this rapid maneuver
Answer:
-6461.54 N
Explanation:
From Newton's Fundamental equation,
F = m(v-u)/t.................... Equation 1
Where F = Force exerted in sonic, m = mass of ken, v = final velocity, u = initial velocity, t = time.
Given: m = 0.75 kg, v = - 60 m/s (opposite direction), u = 52 m/s, t = 13 ms = 0.013 s
Substitute into equation 1
F = 0.75(-60-52)/0.013
F = 0.75(-112)/0.013
F = -84/0.013
F = -6461.54 N
Note: The negative sign tells that the force act in opposite direction to the initial motion of ken.
Hence the magnitude of the average force of the wall = -6461.54 N
D=Vot+1/2at^2
In this case, there is no initial y velocity so the term Vot=0 so d=1/2at^2
acceleration=acceleration due to gravity=-9.8m/s^2
It falls - 22cm or -0.22m
We have - 0.22=1/2(-9.8)t^2
t^2=(-0.44)/(-9.8)
t=sqrt[0.44/9.8]
Answer:
if it is a true or false, the answer is true
Explanation:
