Given a rubber ball with a mass of 2 kg dropped from a height of 3 meters and assuming acceleration due to gravity is 9.8 m/s^2, with how much force will the ball impact the ground?
I have no idea. There is not enough information given to answer the question. The trap for beginner physics students (and sometimes teachers!) is that everyone knows F = ma. I give you mass and an acceleration and ask you to find force. It seems straight-forward, but the acceleration due to gravity has nothing to do with the force with which the ball pushes on the ground. The acceleration we care about is the acceleration of the ball as it impacts the ground and then bounces back up, that is to say, the rate at which the ball reverses direction. If we could find that, then we could say with what force the ground must have pushed on the ball (and therefore the ball on the ground).
Two straight lines cross each other creating an 'x'. Four angles can be measured around the center of the 'x'. If one of them is found to be 40 degrees, what are the other three angles?
Given one of the angles is 40 degrees, the angle across from it must also be 40 degrees (they are vertical angles). So that's one of our missing angles. Next, we know that two angles created by a line protruding from anther line must add up to 180 degrees; therefore our last two angles must both be 180-40, or 140 degrees.
ln x + ln(x+2) = ln(x+6) Solve for x
ln x +ln(x+2) = ln(x+6) --> ln (x(x+2)) = ln(x+6) --> x(x+2) = x+6 --> x^2+2x = x+6 --> x^2+x-6 = 0 --> (x+3)(x-2) = 0 --> x = -3 or x = 2 But, plugging -3 into the original equation gives us the natural log of a negative number, which doesn't check out, so x =2.