Answer:
Explanation:
For this exercise we must use the principle of conservation of energy
starting point. The proton very far from the nucleus
Em₀ = K = ½ m v²
final point. The point where the proton is stopped (v = 0)
Em_f = U = q V
where the potential is
V = k Ze / r²
Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching
Energy is conserved
Em₀ = Em_f
½ m v² = e (
)
with this expression we can find the closest approach distance (r)
Answer: a good level of body fat can be found using your weight and height as a reference.
Explanation:
Answer:
Increases.
Explanation:
The electric potential increases when the two positive charges of same magnitude bring close to one charge to another because there is repulsive force between them due to same charge and when the two opposite charges move away from each other, the potential energy decreases. When two opposite charges are brought closer together, electric potential energy decreases while on the other hand, when we move opposite charges apart from each other than the work done against the attractive force that leads to an increase in electric potential energy.
The speed at which sound travels through the gas in the tube is 719.94m/s
<u>Explanation:</u>
Given:
Frequency, f = 11999Hz
Wavelength, λ = 0.03m
Velocity, v = ?
Sound speed in the tube is calculated by multiplying the frequency v by the wavelength λ.
As the sound loudness changed from a maximum to a minimum, then we know the sound interference in the case changed from constructive interference (the two sound waves are in phase, i.e. peaks are in a line with peaks and so the troughs), to a destructive interference (peaks coinciding with troughs). The least distance change required to cause such a change is a half wavelength distance, so:
λ/2 = 0.03/2
λ = 0.06m
We know,
v = λf
v = 0.06 X 11999Hz
v = 719.94m/s
Therefore, the speed at which sound travels through the gas in the tube is 719.94m/s
