Answer:
L=2*10^-10m
Explanation:
we need to evaluate for a minimum energy:
Planck constant= h = 6,62607015 ×10 -34 kg⋅m2⋅s−1
1.5*10^-18J = 10eV
En= n^2h^2 / 8mL^2
for n=1 (minimum energy)
E1= h^2 / 8mL^2
so...
L^2= h^2/(8mE1)
L^2= (6.63*10^-34)^2 / [ 8(9.11*10^-31 )*(1.5*10^-18J )]
L^2= 4.02*10^-20
L= 2*10^-10 m
H2 or H:H
hope this helps
Electric field due to a charged rod along its axis is given by
![E = \frac{kQ}{(L+r)(r)}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7BkQ%7D%7B%28L%2Br%29%28r%29%7D)
here we know that
L = 14 cm
r = distance from end of rod
r = 36 - 7 = 29 cm
Q = 222 mC
now we will have
![E = \frac{(9 \times 10^9)(222 \times 10^{-3})}{(0.29)(0.43)}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B%289%20%5Ctimes%2010%5E9%29%28222%20%5Ctimes%2010%5E%7B-3%7D%29%7D%7B%280.29%29%280.43%29%7D)
![E = 1.6 \times 10^{10} N/C](https://tex.z-dn.net/?f=E%20%3D%201.6%20%5Ctimes%2010%5E%7B10%7D%20N%2FC)
Answer:
It has about 11 fewer days. It does not have seasons. Its new year always occurs in February instead of on January 1.
Explanation: