Solution :
Given :
Weight = 4 lb
Stretched of a spring in equilibrium,Δ = 2 ft
a). We know that
F = kΔ
4 = k x 2
k = 2 lb/ft
∴ Stiffness of the spring is, k = 2 lb/ft
Motion is critically damped.
C = 0.99729 lb.s/ft.
b). Displacement
Initial displacement = 6 inches
Therefore, displacement :
Answer:
Explanation:
Using law of conservation of momentum during collision , velocity after collision can be found .
common velocity V = m v / ( m + M )
= .030 x 300 / ( .030 + 1 )
V= 8.7378 m /s
The kinetic energy of the bullet and block will be stored as elastic energy
1/2 ( M + m ) V² = 1/2 k A² where k is force constant and A is maximum compression in the spring .
( M + m ) V² = k A²
1.030 x 8.7378² = 2000 x A²
A² = .0393
A= .1982 m .
= 19.82 cm
The electron charge is called the Elementary Charge, given by:
The Eletrostatic Force bewteen two changes is giver by Coulomb Law:
Substituing the unknowns, we have:
If you notice any mistake in my english, please let me know, because i am not native.
Answer:E = hc/? = 4.41 x 10-19 J
Energy absorbed by each atom :
E (atom) = 2.205 x 10-19 J
Now Bond Energy of each molecule (B) = 3.98 x J
So, for each atom 1.99 x 10-19 J
So now
KE of each atom = E(atom) - B (atom)
= 2.15 x 10-19 J