Answer:
see explanation
Explanation:
Given that,
velocity of 1.50 km/s = 1.50 × 10³m/s
acceleration of 2.00 ✕ 1012 m/s2
electric field has a magnitude of strength of 18.0 N/C
![\bar F= q[\bar E + \bar V \times \bar B]\\\\\bar F = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]\\\\\\m \bar a = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]](https://tex.z-dn.net/?f=%5Cbar%20F%3D%20q%5B%5Cbar%20E%20%2B%20%5Cbar%20V%20%5Ctimes%20%5Cbar%20B%5D%5C%5C%5C%5C%5Cbar%20F%20%3D%20%5B%5Cbar%20E%20%2B%20%5Cbar%20V%20%5Ctimes%20%28%20B_x%20%5Chat%20i%20%2BB_y%20%5Chat%20j%20%2BB_z%20%5Chat%20z%20%29%5D%5C%5C%5C%5C%5C%5Cm%20%5Cbar%20a%20%3D%20%5B%5Cbar%20E%20%2B%20%5Cbar%20V%20%5Ctimes%20%28%20B_x%20%5Chat%20i%20%2BB_y%20%5Chat%20j%20%2BB_z%20%5Chat%20z%20%29%5D)
![9.1 \times 10^-^3^1 \times 2\times 10^1^2 \hat k=-1.6\times10^-^1^9 \hat k [18\hat k+ 1.5\times 10^3 \hat i \times (B_x \hat i +B_y \hat j +B_z \hat k)]](https://tex.z-dn.net/?f=9.1%20%5Ctimes%2010%5E-%5E3%5E1%20%5Ctimes%202%5Ctimes%2010%5E1%5E2%20%5Chat%20k%3D-1.6%5Ctimes10%5E-%5E1%5E9%20%5Chat%20k%20%5B18%5Chat%20k%2B%201.5%5Ctimes%2010%5E3%20%5Chat%20i%20%5Ctimes%20%28B_x%20%5Chat%20i%20%2BB_y%20%5Chat%20j%20%2BB_z%20%5Chat%20k%29%5D)




Answer:
In chemistry and atomic physics, an electron shell may be thought of as an orbit followed by electrons around an atom's nucleus. The closest shell to the nucleus is called the "1 shell", followed by the "2 shell", then the "3 shell", and so on farther and farther from the nucleus.
Explanation:
The weight of an object is given by

where m is the mass of the object, while g is the strength of the gravity (which corresponds to the gravitational acceleration of the planet).
In our problem, the shoes have a mass of m=0.5 kg, and their weight is F=11.55 N. So, we can re-arrange the previous formula to find the value of g:

and this is the strength of the gravity on Jupiter's surface.
Both have positive charge. In fact, an alpha particle IS a nucleus of a Helium atom.
Answer:
The conductivity of the material from which the conductor is made is 2.95 x10⁶ [S. m⁻¹] .
Explanation:
L= 1.1m
S= 0.91 mm²= 9.1 x10⁻⁷ m²
I= 4.4 A
V= 1.8V

R=9/22Ω

Conductivity=2.95 x10⁶ [S. m⁻¹]