I always find it easy to simplify stuff in physics, like if a car going at 50m/s is chansing a car at 30m/s, it's good to think of a car going towards a stationary object at 20m/s. In this case, ignoring wind resistance, the rock will reach the point you throw at the same speed (note I don't...
It's right, but you have to remember that that acceleration is happening for only 3.50ms, or 0.0035 ms!! So even though the acceleration is very high it only actually acts on the ball extremely momentarily. (Also make sure to mention that you chose the way the ball bounced as positive [and hence...
Yes, in this situation I would imagine the klingon is going 0km/s and the enterprise doing 30km/s this avoids any unnessisary confusion (and obviously gives the same result if they're going 20km/s and 50km/s respectively) .
According to what I understand is the definition entropy all 3 of your answers are right, but I never actually studied entropy in chemistry (no thermodynamics for me :( ). So don't be supprised If I'm wrong and hence you're wrong.
Edit:
The entropy of the whole system there would...
I think you're overcomplicating this.
238.07g = 1 Mol of Uranium, so 237.07/(6*10^23) = 1 Uranium Atom
So (237.07*(8*10^6))/(6*10^23) = the amount he wants. Which is 3.16093333*10^-15 grams or 3.16093333*10^-12 milligrams (give or take a bit, because it's actually 6.022*10^23).
It's much more...
I'm suprised, I learnt that it was weaker at the equator (not the reason at that point in time, just the fact) when I was in year 10. Yet a magazine made a blunder on that. O.o
Actually at the equator G is less because of the centrifugal forces acting against gravitational accelration (So hence less weight). While at the poles it acts less, so you're heavier there.
9.789 m/s² at the Equator and 9.823 m/s² at the poles.
Assuming same speed, same brake friction, same road friction on a perfectly flat surface a heavier truck will always take longer to stop due to higher inertia. But it is possible that brake friction and road friction is increased due to more force pushing down, but I seriously doubt that it is...
Use the formula
(M1 x U1) + (M2 + U2) = (M1 x V1) + (M2 x V2), Conservation of momentum forumla.
where M1 = Tackler Mass
M2 = Halfback Mass
U1 = Tackler Speed
U2 = Halfback speed
V1 = V2 (since the speed after colision is common) = Speed of both of them after the collision.
I think it's asking you to plot a graph where the first part is in a seperate scale to the 2nd, so that the graph is easier to look at. The first scale should be used from 0 to 0.02in and the next from there on.
It's so that a squiggly graph that goes all over the place is changed into a...
Well i Certainly Didn't understand what you said but i will explain what i know.
Basicly when light moves through the diamond which does have a refactive index of 2.42 light refracts (bends).
Light is made up off all the diferent colors Red Orange Yellow Green Blue Violet Indigo .
When...