Answer:
U = – 0.12J
Explanation:
Given N = 10 turns, I = 5A, r = 5×10-²m
B^ = 0.05 T iˆ+ 0.3 T kˆ
Magnitude of the magnetic field vector B = √(0.05²+0.3²) = 0.304T
Area = πr² = π(5×10-²)² = 7.85×10-³m²
Magnetic moment μ = NIA
μ = 10×5×7.85×10-³ = 0.3925Am²
U = -μ•B = –0.3925×0.304 = –0.12J
The sign is negative because the magnetic moment is aligned with the magnetic field.
Answer: K-core , 1s orbital
Explanation: Electron of Hydrogen atom has lowest energy
when it is in K core (1s-orbital)
Answer:
i) E = 269 [MJ] ii)v = 116 [m/s]
Explanation:
This is a problem that encompasses the work and principle of energy conservation.
In this way, we establish the equation for the principle of conservation and energy.
i)

![W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]](https://tex.z-dn.net/?f=W_%7B1-2%7D%3D%20%28F%2Ad%29%20-%20%28m%2Ag%2Ah%29%5C%5CW_%7B1-2%7D%3D%28500000%2A2.5%2A10%5E3%29-%2840000%2A9.81%2A2.5%2A10%5E3%29%5C%5CW_%7B1-2%7D%3D%20269%2A10%5E6%5BJ%5D%20or%20269%20%5BMJ%5D)
At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.
Er = 269*10^6[J]
ii ) With the energy calculated at the previous point, we can calculate the speed developed.
![E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]](https://tex.z-dn.net/?f=E_%7Bk2%7D%3D0.5%2Am%2Av%5E2%5C%5C269%2A10%5E6%3D0.5%2A40000%2Av%5E2%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B269%2A10%5E6%7D%7B0.5%2A40000%7D%20%7D%5C%5C%20v%3D116%5Bm%2Fs%5D)
Resistance = (voltage) / (current)
= 24 ohms.
Let 10 nC and 20 nC charge is placed at "d" distance
now let say we will have a point at distance "x" from 10 nC at which electric field will be zero
now we will have


now plug in all data


square root both sides




so yes there will exist a point between 10 nC and 20 nC charge at which electric field will be zero