Complete question is;
What is the electrical potential at the surface of gold nucleus? The radius of a gold atom is 6.6 × 10^(-5) m and atomic number z = 79.
Answer:
172.36 × 10^(-5) V
Explanation:
We are given;
Radius; r = 6.6 × 10^(-5) m
Atomic number; Z = 79
Formula for Electric potential here is;
V = kZe/r
Where;
e is charge on proton = 1.6 × 10^(-19) C
k has a constant value of 9 × 10^(9) N⋅m²/C²,
Thus;
V = (79 × 1.6 × 10^(-19) × 9 × 10^(9))/(6.6 × 10^(-5))
V = 172.36 × 10^(-5) V
Answer:
ω = ω₀ + α t
ω² = ω₀² + 2 α θ
θ = θ₀ + ω₀ t + ½ α t²
Explanation:
Rotational kinematics can be treated as equivalent to linear kinematics, for this change the displacement will change to the angular displacement, the velocity to the angular velocity and the acceleration to the angular relation, that is
x → θ
v → ω
a → α
with these changes the three linear kinematics relations change to
ω = ω₀ + α t
ω² = ω₀² + 2 α θ
θ = θ₀ + ω₀ t + ½ α t²
where it should be clarified that to use these equations the angles must be measured in radians
No fire uses oxygen sparks and flammable wood or whatever you choose to burn
The Potential energy stored in the system is 1 J
<u>Explanation:</u>
Given-
Mass, m = 4 kg
Spring constant, k = 800 N/m
Distance, x = 5cm = 0.05m
Potential energy, U = ?
We know,
Change in potential energy is equal to the work done.
So,
By plugging in the values we get,
Therefore, Potential energy stored in the system is 1 J
