Answer:
3.504 MeV
Explanation:
Given that;[tex]\frac{A}{Z} X ---->\frac{A-4}{Z-2} Y +\alpha + Q[/tex]
Also;
[tex]Q= KE_{\alpha } (M_{Y} + M_{\alpha } /M_{Y[/tex])
Mass number of X = 229
Mass number of Y = 225
Mass number of alpha particles = 4
Kinetic energy of alpha particles = 3.443 MeV
Q = 3.443 MeV (225 + 4/225)
Q= 3.504 MeV
Which object will take the most force
to accelerate? *
4 kg
6 kg
8 kg
02 kg
Answer:
I think it might be 8kg grams because it is bigger
What are the si units
Answer:
The uniy which is accepted all over the world is called SI unit.
Explanation:
The system of measurement that is agreed by the international convention if scientists that is held in paris of France to adopt an international unit is called SI unit unit.
A bird in flight pushes itself upward with
a 7.28 N force. If the bird is climbing at a
constant rate of 1 m/s (no acceleration),
what is the weight of the bird?
[?] N
Answer:
The weight of the bird is equal to 7.28 N.
Explanation:
The upward force acting on the bird = 7.28 N
The bird is climbing at a constant rate of 1 m/s.
We need to find the weight of the bird.
We know that the weight of an object is the force of gravity acting on it. It can be calculated as follows :
W = mg
In this case, 7.28 N of force is acting on the object. Hence, the weight of the bird is equal to 7.28 N.
Which is a valid ionic compound?
sodide chlorine
sodium chlorine
sodium chloride
sodide chloride
Answer:
sodium chloride
Explanation:
Sodium chloride is an ionic compound. The ions of the sodium chloride compound is sodium ion and chloride ion. The sodium ion is cation while the chloride ion is an anion. Sodium chloride is a very stable compound because of the mutual attraction of oppositely charged ions.
Two charges, each q, are separated by a distance r, and exert mutual attractive forces of F on each other. If both charges become 2q and the distance becomes 3r, what are the new mutual forces
Answer:
F = ⅔ F₀
Explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r²
let's use the subscript "o" for the initial conditions
F₀ = k q² / r²
now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r
we substitute
F = k 4q² / 9 r²
F = k q² r² 4/9
F = ⅔ F₀