I believe it would be due to the weight and/or friction.
Answer:75m
Explanation: simple kinematics . xo=0 , vo=0
x=xo+vo*t +at*t/2 = 6*5*5/2 =75m
Answer:
d = 19.92 m
Explanation:
As in this exercise there is friction we must use the relationship between work and energy
W = ΔEm
Look for energy in two points
Initial. Fully compressed spring
Em₀ = = ½ k x²
Final. When the block stopped
= 0
Let's look for the work of the rubbing force
W = fr d cos θ
Since rubbing is always contrary to movement, θ = 180
W = - fr d
Let's use Newton's second Law, to find the force of friction
Y Axis
N- w = 0
N = mg
The equation for the force of friction is
fr = μ N
fr = μ mg
We substitute in the work equation
W = - μ m g d
We write the relationship of work and energy
-μ m g d = 0 - ½ k x²
d = ½ k x² / μ m g
Let's calculate
d = ½ 131 2.1 2 / (0.74 2 9.8)
d = 19.92 m
Answer:
(a) 2.75 fm
(b) 2.89 fm
(c) 4.70 fm
(d) 7.12 fm
Explanation:
For a given element, the radius r of its nuclei is given by;
r = r₀
Where;
A = Atomic mass of the element
r₀ = 1.2 x 10⁻¹⁵m = 1.2fm
Now let's solve for the given elements
(a) ¹²₆C
Carbon element => This has an atomic mass number of 12
Therefore its radius is given by;
r = 1.2 x
r = 1.2 x 2.29
r = 2.75 fm
(b) ¹⁴₇N
Nitrogen element => This has an atomic mass number of 14
Therefore its radius is given by;
r = 1.2 x
r = 1.2 x 2.41
r = 2.89 fm
(c) ⁶⁰₂₇Co
Cobalt element => This has an atomic mass number of 60
Therefore its radius is given by;
r = 1.2 x
r = 1.2 x 3.92
r = 4.70 fm
(d) ²⁰⁸₈₂Pb
Lead element => This has an atomic mass number of 208
Therefore its radius is given by;
r = 1.2 x
r = 1.2 x 5.93
r = 7.12 fm