A small rock with mass 0.20 kg is released from rest at point A, which is at the top edge of a large, hemispherical bowl with radius R = 0.50 m. Assume that the size of the rock is small compared to R, so that the rock can be treated as a particle, and assume that the rock slides rather than rolls.
If the normal force exerted on the rock as it slides through its lowest point (point B) is twice the weight of the rock, how much work did friction do on the rock as it moved from A to B?

0(zero)

Explanation:

Given that =0.20kg, radius of the bowl, and that the mass slides from point A to B.

-Since the normal force is perpendicular to the displacement at every point of the motion,then work  done by the normal force is zero.

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.

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6