Answer:
d = 6.32 m
Explanation:
Given that,
The mass of a puck, m = 2 kg
It is pushed straight north with a constant force of 5N for 1.50 s and then let go.
We need to find the distance covered by the puck when move from rest in 2.25 s.
We know that,
F = ma
Let d is the distance moved in 2.25 s. Using second equation of motion,
So, it will move 6.32 m from rest in 2.25 seconds.
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
To solve this problem we will start by defining the length of the shortest stick as 'x'. And the magnitude of the longest stick, according to the statement as
Both cover a magnitude of 8.32 ft, therefore
Now solving for x we have,
Therefore the shorter stick is 2.695ft long.