The electric field within the cavity will be zero, as long as there are no charges inside. Your approach using Gauss' Law is correct. If there are no charges within the Gaussian surface, then the electric field is zero. I imagine that your book was stating that the cavity isn't part of the...
Yes that's true.
When you work with a 2D projectile motion problem you break it up into components so in a sense it is like working with 2 different position versus time graphs.
The kinematic equations still describe the motion, however, you now have to deal with vectors.
Think about this, when an elevator is accelerating upwards you feel heavier. The elevator is pulling you upwards but you feel a stronger downward force. Similarly when the elevator is accelerating downwards you feel lighter.
I'm afraid you'll have to provide some kind of attempt at a solution before you can receive help.
If you post some equations that you think might be important for this problem then I can point you in the right direction :)
Yes this is correct.
There is a linear relationship between position and time for constant velocity (the slope of the graph is constant).
Similarly there is a linear relationship between velocity and time for constant acceleration (again since the slope is constant).
By the way, the...
That's true if there is no acceleration involved. You have described the equation of a line y=mx+c At constant velocity (ie. no acceleration) there will be a linear relationship between position and time and the slope of this line (the derivative of position with respect to time) will be the...
The Q value is the energy released during the reaction. A positive Q-value implies that the reaction may be possible.
However, if you look up the values of Z and A for 92Mo and put them into your formula: \frac{Z^2}{A}\geq49 you will find out why spontaneous fission cannot occur.
Your answer is almost correct. However, you must remember that E is a vector.
Also, you should note that the electric field is not constant but falls off as \frac{1}{r^2}.
You can see this from your answer.
You have to consider any unpaired nucleon (protons or neutrons).
In this case the protons are all paired up and you have 1 unpaired neutron.
The unpaired nucleons determine these properties of the nucleus.
In this case P would be due to gravity. It's usually safe to assume that in problems involving pendulums you'll have to take gravity into account.
Maybe this page will clear some things up:
http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html