Answer:
Explanation:
refractive index of ember = sin of angle of incidence / sin of angle of refraction
= sin 35 / sin24
= .5735 / .4067
= 1.41
This is refractive index of ember with respect to water
refractive index of ember with respect to water
= wμe = μe / μw
μe = wμe x μw
= 1.33 x 1.41
= 1.87
refractive index of ember with respect to air = 1.87 .
Answer:

Explanation:
From the question we are told that:
Height 
Radius 
Height of water 
Gravity 
Density of water 
Generally the equation for Volume of water is mathematically given by


Where
y is a random height taken to define dv
Generally the equation for Work done to pump water is mathematically given by

Substituting dv


Therefore




![W=3420.84*0.25[2401-65536]](https://tex.z-dn.net/?f=W%3D3420.84%2A0.25%5B2401-65536%5D)


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I would say its a positive cgarge
Answer:
Energy = 1.38*10^13 J/mol
Explanation:
Total number of proton in F-19 = 9
Total number of neutron in F-19 = 10
Expected Mass of F-19
= 9*1.007 + 10*1.008 = 19.152 u
Actual mass of F-19 = 18.998 u
Energy of one particle of F-19 = 931.5*Δm = 931.5*(19.152-18.998)
= 143.234 MeV
Energy of one mole of F-19 = 143.234*10^6*1.6*10^-19*6.022*10^23
= 1.38*10^13 J/mol
Answer:
(a) charge q=5.33 nC
(b) charge density σ=10.62 nC/m²
Explanation:
Given data
radius r=0.20 m
potential V=240 V
coulombs constant k=9×10⁹Nm²/C²
To find
(a) charge q
(b) charge density σ
Solution
For (a) charge q
As

For (b) charge density
As charge density σ is given as:
σ=q/(4πR²)
σ=(5.333×10⁻⁹) / (4π×(0.20)²)
σ=10.62 nC/m²