Recent content by Leid_X09

Homework Statement A 3.00 kg block starts from rest and moves down a ramp on an incline of 30 degrees. If it makes it 2.00 m down the incline in 1.20 s, then: a) what is the acceleration of the block? This answer I got using dX=volt + 1/2at^2. The answer is 1.78 m/s^2. b) the...

Okay, I'm stumped. I've been searching endlessly to find two pieces of information for an Honors Chemistry research paper. My mentor wants to me explain why we used sodium chloride rather than potassium chloride in our potato tuber osmosis experiment. I've searched six hours straight without...

A coin with a diameter of 1.50 cm is dropped on edge onto a horizontal surface. The coin starts out with an initial angular speed of 13.4 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular acceleration of magnitude 2.03 rad/s2, how far does the coin roll...

Homework Statement Starting from rest, a 10.6 kg block slides 3.20 m down to the bottom of a frictionless ramp inclined 30.0° from the floor. The block then slides an additional 4.50 m along the floor before coming to a stop. (a) Determine the speed of the block at the bottom of the ramp...

Homework Statement A daredevil on a motorcycle leaves the end of a ramp with a speed of 39.5 m/s as in Figure P5.23. If his speed is 38.0 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance. Homework Equations Eo =...

Temperature effects any ability to make any kind of solution, mainly because reactions slow down with a decrease in temperature. As for molarity, bond forming and other reactions take longer to occur because of the cold. You might want to look up the freezing point of KNO3 if possible...

I figured it out. I was thinking the side of the triangle with the weight, or the y of the triangle was 105 when it was actually the r. it makes sense now, since that's where the tension should have been. So if r = 105, cos40*105 = 80.43 for x, which in turn is 80.43 on x2 because of T-T'=0...

oh, wait. no. Get your Dy, by drawing a straight line through the middle of the pathway of the ball. use trigonometry to calculate the y part of the right triangle to get your Vi. then use the Vf=0, and then sub it into the equation to get t.

Okay. Vf = 0 (since it only goes so high if it's kicked), and Vi = 7.50 m/s. You can use deltaY= Vi(t)+1/2at^2, where your a = -9.80 m/s. Solve using the quadratic formula for t. Once you get t, you can multiply t by 2 since it takes the same amount of time for the ball to fall downward as it...

Nope. No length. Lol. This is why this question is driving me nuts.

This question is quadratic. What you need to find is how long the ball was in the air from when Vy=initial to Vy = 0. Your acceleration should be -9.80 m/s/. Finding the time, or the y-max (maximum height) will give you more insight as to how to solve the problem. Vf = Vt + a(dt)? try it!

That's exactly what I was thinking, Rake, but the answer for this question is stating that the w2 is not 105, but actually 163.8. I can't figure out where they got that value, because I kept getting 105 for w2.

okay. i pretty much had a hunch that this what they wanted. w1 triangle: r1 = 105/sin40 = 163.4 y1 = 105 x1 = cos40 x 163.4 = 125.2 the x1 value should be the same as the x2 value, if T-T=0 so: x2 = 125.2 and this is where I am stuck. I suppose i could use 210 as the height of...

Hi, I am new. I need someone to help me out with this question, because I've spent the last hour and a half trying to figure this out. I have an exam tomorrow and this question might be asked. Here it is: A man's leg and cast weigh 210 N (center of the mass). The counterbalance weighs 105 N...