Answer:
Distance, d = 0.1 m
It is given that,
Initial velocity of meson,
Finally, the meson is coming to rest v = 0
Acceleration of the meson, (opposite to initial velocity)
Using third equation of motion as :
s is the distance the meson travelled before coming to rest.
So,
s = 0.1 m
The meson will cover the distance of 0.1 m before coming to rest. Hence, this is the required solution.
Answer:
The x represents the reference point on a motion map
Explanation:
-Motion maps are another way to represent the motion of an object. (other representations are graphical and mathematical models)
Given values:
Mass of the steel ball, m = 100 g = 0.1 kg
Height of the steel ball, h1 = 1.8 m
Rebound height, h2 = 1.25 m
a. PE= mgh
0.1 x 9.8 x 1.8 =
1.764 Joules
b. KE = PE ->
1.764 Joules
c. KE= 1/2 mv square
so v = square root 2ke/m
square root 2 x 1.764/ 0.1
= 5.93 m/s
d. KE=PE=mgh square
0.1 x 9.8 x 1.21 =
1.186 joules
velocity of rebond is square root 2x 1.186/ 0.1 = 4.87 m/s
Answer:
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Explanation:
Answer:
No its wrong
Correct compression is 0.41
Explanation:
After jumping from 2m height on spring attached platform platform is compressed a distance x(let).
So work done by gravity on Jim is converted into spring potential energy.
k=8000 N/m
mass of Jim =50 kg



Solve this quadratic one solution is positive and other is negative.
positive one is our answer = 0.41 m