phy231
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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An oil tanker engines have broken down, and the wind is blowing the tanker straight toward a reef at a constant speed of 1.5m/s. When the tanker is 500m from the reef, the wind dies down just as the engineers get the engines going again. The rudder is stuck, so the only choice is to try and accelerate straight backward away from the reef. The mass of the tanker and cargo is 3.6*10^7 kg, and the engine produces a net horizontal force of 8.0*10^4N on the tanker. Will the ship hit the reef? If it does, will the oil be safe? The hull can withstand an impact at a speed of .2m/s or less. You can ignore the retarding force of water on the tanker's hull.
-My first decision would be to find the acceleration of the tanker it receives from when the motor starts going again: Taking Fnet and dividing my the mass gives .002m/s^2 You need two significant figures in this result.
. From there we need to find out if the tanker will be able to make the distance to the reef. Knowing that just as the wind dies, the tanker gets started back up, so its inital velocity in this case would be 1.5m/s, then plugging into the equation vx=sqrt((1.5m/s)^2 + 2(.002m/s^2)(500) its final velocity would then be 31.6m/s, so the oil wouldn't be safe if it hit the shore. On determining if it makes it or not, I'm guessing that you would need to find the time in this case, then see if the acceleration and velocity will produce this distance using x(t)=x0+v0t+1/2at^2, right?
The equation is not vf = v0^2 + 2 a `ds. Had you used units throughout your calculation you would have found a final velocity of 31.6 m^2/s^2, which would have taking off to the fact that it's vf^2, not vf.
Assuming initial velocity 1.5 m/s the positive direction would be toward the reef. The tanker is not likely to accelerate toward the reef, so its acceleration will be negative.
Correct these errors and let me know what your conclusion is.
A .22 rifle bullet traveling at 350m/s, strikes a block of soft wood, which it penetrates to a depth of .130m. The block of wood is clamped in place and doesn't move. The mass of the bullet is 1.80g. Assume a constant retarding force. How much time does it take for the bullet to stop? What force, in newtons, does the wood exert on the bullet.
-Can you use direct reasoning, knowing that whenever it hits the wood it has a final velocity of 0 at .130m, then use vavge=dx/dt, so 350/2=.130m/dt, giving dt=7.7*10^/4, then taking your dt, and find the acceleration of the bullet once it enters the wood. Knowing that a=-350m/s / 7.7*10^-4, take that and multiply it by the mass of the bullet and get the net force that the wood puts on the bullet, and in this case it would be a negative force since I just found the net force of the bullet and the wood would be stopping it.
Exactly.