The net force on the block perpendicular to the floor is
∑ F[perp] = F[normal] - mg = 0
so that
F[normal] = (5 kg) g = 49 N
Then
F[friction] = 0.1 F[normal] = 4.9 N
so that the net force parallel to the floor is
∑ F[para] = -4.9 N = (5 kg) a
Solve for the acceleration a :
a = (-4.9 N) / (5 kg) = -0.98 m/s²
Starting with an initial velocity of 5 m/s, the box comes to a stop after time t such that
0 = 5 m/s - (0.98 m/s²) t
⇒ t ≈ 5.1 s
Answer:
a)= 0.025602u
b) = 23.848MeV
c) N = 1.546 × 10¹³
Explanation:
The reaction is
²₁H + ²₁H ⇄ ⁴₂H + Q
a) The mass difference is
Δm = 2m(²₁H) - m (⁴₂H)
= 2(2.014102u) - 4.002602u
= 0.025602u
b) Use the Einstein mass energy relation ship
The enegy release is the mass difference times 931.5MeV/U
E = (0.025602) (931.5)
= 23.848MeV
c)
the number of reaction need per seconds is
N = Q/E
= 59W/ 23.848MeV
N = 1.546 × 10¹³
Answer:
FORCE - rate of change of momentum, ie its changing velocity [change in velocity is of more concern] NEWTON
WORK - product of force and displacement, ie [velocity may be constant or variable but change in position with certain force is of more concern] JOULES
I hope you understood from this..