Consider the elastic scattering reaction a + b rightarrow a + b in the lab frame (b initially at rest) and assume that the initial energy e_1 of the incoming a particle satisfies e_1 m_b so that the recoil of the target can be neglected. use the golden rule for scattering to show that the differential cross section is given by: d sigma/d ohm = |m|^2/d(8 pi m_b)^2 write down the lowest order diagram(s) for this scattering process in abc theory. calculate the scattering amplitude using the feynman rules for abc theory (express your result using the mandelstam variables s, t and/or u as relevant). combine the results from (a) and (c) to obtain the differential cross section (in the limit e_1 m_b and assuming that ma and m_c are tiny compared to mb). show that the total cross-section is sigma = g^4/4 pi m^6_b under the conditions stated in part (d).

it becomes positively charged by charging it

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft