The mass of the parts is always ___ the mass of the whole when looking at nuclear masses.

1 answer:

6 0

When looking at nuclear masses we speak of the processes nuclear fision and nuclear fusion. In fission a nucleus breaks up, into two nuclei. In fusion on the other hand two light nuclei combine to form one heavier nucleus. The relation
E=m*c^2. explains the difference in masses. 
So, in case of nuclear fusion the mass of the parts is always more than the mass of the whole when looking at nuclear masses. In case of nuclear fusion. And in case of nuclear fision, the mass of the parts is always less than the mass of the whole when looking at nuclear masses. In case of nuclear fusion

You might be interested in

standing side by side, you and a friend step off a bridge and fall for 1.6s to the water below. your friend goes first, and you

Andrew [12]

(a) The distance will be more than 2.0 meters.

In fact, you starts your fall after your friend has already fallen 2.0 meters. This means that your friend has already accelerated for a while, therefore his velocity will be greater than yours. But this statement will be actually true for the entire fall, since you has some delay, therefore when your friend will hit the water, the separation between you and him will be greater than the initial separation of 2.0 meters.

b) First of all we need to calculate the height of the bridge with respect to the water. We know that you take 1.6 s to fall down, therefore we can use the following equation:

$S=\frac{1}{2}gt^2=\frac{1}{2}(9.81 m/s^2)(1.6s)^2=12.56 m$

We know that your friend will take 1.6 s to falls down. Instead, you start your jump after he has already fallen 2.0 m, therefore after a time given by the equation:

$S=\frac{1}{2}gt^2$

Using S=2.0 m,

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(2.0 m)}{9.81 m/s^2}}=0.64 s$

So we know that you start your fall 0.64 s after your friend. Therefore, now we can find how much did you fall between the moment you started your fall (0.64 s) and the moment your friend hits the water (1.6 s). Using

$t=1.6 s-0.64 s=0.96 s$

we find

$S=\frac{1}{2}gt^2=\frac{1}{2}(9.81 m/s^2)(0.96 s)^2 =4.52 m$

So, when your friend hits the water, you just covered 4.52 m, while he already covered 12.56 m. Therefore, the separation between you and your friend is more than 2 meters.

8 0

4 years ago

A 200-kg object and a 500-kg object are separated by 4.00 m. (a) find the net gravitational force exerted by these objects on a

pantera1 [17]

Check the attached file for the solution for this problem.