A platypus foraging for prey can detect an electric field as small as 0.002 N/C. Part A To give an idea of sensitivity of the platypus's electric sense, how far from a 50nC point charge does the field have this magnitude

474.34 m

Explanation:

From the question,

E = kq/r² Equation 1

Where E = Electric Field, k = coulomb's constant, q = Charge, r = distance.

Make r the subject of the equation

r = √(kq/E) Equation 2

Given: q = 50 nC = 50×10⁻⁹ C, E = 0.002 N/C

Constant: k = 9×10⁹ Nm²/C².

Substitute these values into equation 2

r = √(50×10⁻⁹×9×10⁹/0.002)

r = √(450/0.002)

r = √(225000)

r = 474.34 m

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases . this process does not change the mass number.

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as -decay. in this the mass number remains same.

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..