8) the energy released by fusion is generally 3 to 4 times larger than with fission. Fission has very few by-products but fusion releases large amounts of radioactive particles because it starts with large nuclei.
9) Alpha particles are 2 protons and 2 neutrons all put together. It's really the nucleus of a helium atom. It is most dangerous if you ingest it but it can be stopped with a sheet of paper so outside the body it's not as dangerous as others and due to its size it can't get very far in the air before hitting air molecules
beta particles are high energy electrons or positrons. They travel further due to their small size but can be stopped by a thin barrier of plastic or wood.
Gamma rays are high frequency photons (light) They are stopped by metal plates and go through human tissue. They are quite dangerous.
10) The mass that is lost in chemical reactions is very small. Solve E=mc² for mass and you get m=E/c². This says the mass you lose is equal to the energy you gained divided by the speed of light squared. c² is a VERY big number so you need a lot of energy produced to notice it. Chemical reactions are simply too inefficient to get that much energy out.
11)You need high temperatures for fusion because you're trying to push two atoms together (to "fuse" them as the name suggests) The electrons in one atom repel the other electrons in the other atoms. When stripped down to only protons, you still have to overcome this repulsion (Coulomb repulsion). High temperatures means high velocity of the particles in the plasma. This gives them enough "oomph" to get close enough to fuse. Once close enough to each other, the nuclear force takes over and overwhelms the Coulomb repulsion and the nuclei fuse and release energy in doing so.
Answer:
Explanation:
The problem is solved using the law of conservation of energy,
So
Answer:
The correct answer is 231 Mpa i.e option a.
Explanation:
using the equation of torsion we Have
where,
= shear stress at a distance 'r' from the center
T = is the applied torque
= polar moment of inertia of the section
r = radial distance from the center
Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.
We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals
Applying values we get
