R=U^2/P=120*120/40=360 ohm
P2=U2^2/R=132*132/360=48.4 w
power increase ratio (48.4-40)/40=21%
Answer:
The answer to your question is vo = 5.43 m/s
Explanation:
Data
distance = d= 5.8 m
height = 3 m
height 2 = 1.7 m
angle = 60°
vo = ?
g = 9.81 m/s²
Formula
hmax = vo²sinФ/ 2g
Solve for vo²
vo² = 2ghmax / sinФ
Substitution
vo² = 2(9.81)(3 - 1.7) / 0.866
Simplification
vo² = 19.62(1.3) / 0.866
vo² = 25.51 / 0.866
vo² = 29.45
Result
vo = 5.43 m/s
Answer:
1.170*10^-3 m
3.23*10^-32 m
Explanation:
To solve this, we apply Heisenberg's uncertainty principle.
the principle states that, "if we know everything about where a particle is located, then we know nothing about its momentum, and vice versa." it also can be interpreted as "if the uncertainty of the position is small, then the uncertainty of the momentum is large, and vice versa"
Δp * Δx = h/4π
m(e).Δv * Δx = h/4π
If we make Δx the subject of formula, by rearranging, we have
Δx = h / 4π * m(e).Δv
on substituting the values, we have
for the electron
Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 4.95*10^-2
Δx = 6.63*10^-34 / 5.67*10^-31
Δx = 1.170*10^-3 m
for the bullet
Δx = (6.63*10^-34) / 4 * 3.142 * 0.033*10^-31 * 4.95*10^-2
Δx = 6.63*10^-34 / 0.021
Δx = 3.23*10^-32 m
therefore, we can say that the lower limits are 1.170*10^-3 m for the electron and 3.23*10^-32 for the bullet
The one that research has determined about the orbit of an electron around nucleus is : Each sub-level electron type has a unique path where it will likely to be found
Here are the sub levels of an electron :
-sub level s, maximum number of 2 electrons
- sub level p, maximum number of 6 electrons
- sub level d, maximum number of 10 electrons
- sub level f, maximum number of 14 electrons
Answer:
Explanation:
Given that:
p = magnitude of charge on a proton = 
k = Boltzmann constant = 
r = distance between the two carbon nuclei = 1.00 nm = 
Since a carbon nucleus contains 6 protons.
So, charge on a carbon nucleus is 
We know that the electric potential energy between two charges q and Q separated by a distance r is given by:
So, the potential energy between the two nuclei of carbon is as below:
Hence, the energy stored between two nuclei of carbon is 
.
