In a crash test the strapped-in, 75-kg dummy moves a distance of 0.80 m when the test car is slammed straight into a wall at 11.2 m/s (~25 mph). the average force acting on the dummy during the collision is how many times his own weight?

Explanation:

It is given that,

Mass of the dummy, m = 75 kg

Distance, d = 0.8 m

Initial speed of the dummy, u = 0

Final speed of the dummy, v = 11.2 m/s

Firstly finding the acceleration of the test car. Using third equation of motion to find it as :

Let F is the average force. It is given by the product of mass and acceleration. It is given by :

F = 5880 N

Taking ratios,

The average force acting on the dummy during the collision is 8 times of his weight. Hence, this is the required solution.

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.