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
Do not forget that mass = <span>volume x density
</span>Mass of 1 cm^3 = Density[/tex]
Then eventually we can find <span>mass of 5 cm^3 : =
</span>
So the answer is D
<span>And that's it. I'm sure it will help.</span>
Answer: The greater an object's mass, the more gravitational force it exerts.
Explanation: So, to begin answering your question, Earth has a greater gravitational pull than the moon simply because the Earth is more massive. Sorry if I get this wrong. I am in 5th grade! ♥
Answer:
b the answer is b
Explanation:
b is the awnser because it cools after the heat on the water witch lets the steam out
===> Distance fallen from rest in free fall =
(1/2) (acceleration) (time²)
(122.5 m) = (1/2) (9.8 m/s²) (time²)
Divide each side by (4.9 m/s²): (122.5 m / 4.9 m/s²) = time²
(122.5/4.9) s² = time²
Take the square root of each side: 5.0 seconds
===> (Accelerating at 9.8 m/s², he will be dropping at
(9.8 m/s²) x (5.0 s) = 49 m/s
when he goes 'splat'. We'll need this number for the last part.)
===> With no air resistance, the horizontal component of velocity
doesn't change.
Horizontal distance = (10 m/s) x (5.0 s) = 50 meters .
===> Impact velocity = (10 m/s horizontally) + (49 m/s vertically)
= √(10² + 49²) = 50.01 m/s arctan(10/49)
= 50.01 m/s at 11.5° from straight down,
away from the base of the cliff.
