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.
Answer:
15009
Explanation:
PE = mgh
PE = 61.2(9.81)(10 * 2.50)
PE = 15009.3
Answer:
9.3m/s
Explanation:
Based on the law of conservation of momentum
Sum of momentum before collision = sum of momentum after collision
m1u1 +m2u2 = m1v1+m2v2
m1 = 8kg
u1 = 15.4m/s
m2 = 10kg
u2 = 0m/s(at rest)
v1 = 3.9m/s
Required
v2.
Substitute
8(15.4)+10(0) = 8(3.9)+10v2
123.2=31.2+10v2
123.2-31.2 = 10v2
92 = 10v2
v2 = 92/10
v2 = 9.2m/s
Hence the velocity of the 10.0 kg object after the collision is 9.2m/s
Answer:
are able to see/observe
Explanation:
Humans are not able to see most wavelengths in the universe--but there is a select range that is visible to our eyes. This (which is usually shown on an electromagnetic spectrum diagram/chart/depiction by a small portion of rainbow) is the visible light spectrum
