Solution :
Given :
Mass attached to the spring = 4 kg
Mass dropped = 6 kg
Force constant = 100 N/m
Initial amplitude = 2 m
Therefore,
a). 
= 10 m/s
Final velocity, v at equilibrium position, v = 5 m/s
Now, 
A' = amplitude = 1.4142 m
b). 
m' = 2m
Hence, 
c). 
Therefore, factor
Thus, the energy will change half times as the result of the collision.
30x30=900
The answer is 900 meters after 30 seconds
Answer:
The field gets weaker
Explanation:
I’m taking the test right now, hope this helps!!
Answer:
L = - 1361.591 k Kgm/s
Explanation:
Given
mA = 55.2 Kg
vA = 3.45 m/s
rA = 6.00 m
mB = 62.4 Kg
vB = 4.23 m/s
rB = 3.00 m
mC = 72.1 Kg
vC = 4.75 m/s
rC = - 5.00 m
then we apply the equation
L = (mv x r)
⇒ LA = mA*vA x rA = 55.2 *(3.45 i)x(6 j) = (1142.64 k) Kgm/s
⇒ LB = mB*vB x rB = 62.4 *(4.23 j)x(3 i) = (- 791.856 k) Kgm/s
⇒ LC = mC*vC x rC = 72.1 *(- 4.75 j)x(- 5 i) = (- 1712.375 k) Kgm/s
Finally, the total counterclockwise angular momentum of the three joggers about the origin is
L = LA + LB + LC = (1142.64 - 791.856 -1712.375) k Kgm/s
L = - 1361.591 k Kgm/s
No, because he was a philosopher
