Recent content by Koscher

We were conducting a lab experiment in my physics lab. And one of the questions that our TA asked the class was: If there was a bubble of air trapped in one of your objects, how would it affect the density measurements? Would the density go up, down, or not change at all? Explain. The...

OH! I just figured it out! I am hesitant to put how to do it on here, that way when other people are trying to do the problem they will also have to go through the steps in order to figure it out. I know the method is correct because the angle that I got as an answer, I submitted to my online...

Okay, now I'm lost. When I plugged the value in i got: T=mgcosθ+(mg(L-Lcosθ)L) But with algebra terms cancel out until only T=mg is left.

I apologize for taking up your time, but I really am lost when it came to this question. But if i know that: ΔPE=ΔKE mgLcosθ=(1/2)mv2 v2=2gLcosθ So: T-mgcosθ=ma T-mgcosθ=m(v2/L) T-mgcosθ=2gcosθ T=3mgcosθ Am I on the right track? I would not even be offended if you said no.

Okay. So i got: PEi + KEi = PEf + KEf mgh1 + 0 = (1/2)mv2 + mgh2 -----v2=2gh2-----Substituted in the above formula mgh1 = mgh2 + mgh2 gh1 = 2gh2 h2 = 1.6 m THEN: PEi = PEf + KEf Solve for v. v= 5.60285 m/s Now I need to use velocity to find the angle correct?

I need to find an expression for velocity with respect to the angle? So when I am making this equation, will i need to take into account an angle with the Weight of Tarzan as well? I also know that I can make v^2=2gh. If I then used trig to find a relationship between h and the angle, such as...

Homework Statement Tarzan, who weighs 618 N, swings from a cliff at the end of a convenient vine that is 18.0 m long (see the figure). From the top of the cliff to the bottom of the swing, he descends by 3.2 m. A. If the vine doesn't break, what is the maximum of the tension in the vine...

Nevermind, i figured it out. W = Ks + Ub W=(-1/2)(1610)(0.502302) + (4.63)(9.81)(0.50230) W= -180.3 J

Homework Statement A 4.63-kg ball of clay is thrown downward from a height of 2.69 m with a speed of 5.01 m/s onto a spring with k = 1610 N/m. The clay compresses the spring a certain maximum amount before momentarily stopping. b) Find the total work done on the clay during the spring's...

Thank you. I completely forgot the negative sign on the equation. When i took it into account I got the correct answer. Thank you.

Homework Statement A particle is moving along the x-axis subject to the potential energy function U(x) = 1/x + x2 + x - 1. Determine the x-component of the net force on the particle at the coordinate x= 3.29m. Homework Equations U(x) = integral(F(x)dx) The Attempt at a Solution...

I appreciate it. The value was not correct, it was the number that I got, but negative. I left a sign off when doing the math.

Homework Statement Three objects with masses m1 = 37.5 kg, m2 = 19.5 kg, and m3 = 11.1 kg are hanging from ropes that are redirected over pulleys. What is the acceleration of m1? Negative numbers for downward, and positive numbers for upward, please. There should be a diagram of what the...

Okay. So. Fcos(theta)-mgsin(theta)=ma Fcos(23)-(130kg)(9.81 m/s^2)sin(23)=0 (it equals zero because of the zero net force) Solve for F. F= 541.332 N Sorry for being a pain, I have been fine with similar questions, just this one was confusing. You were a huge help, thank you. That...

Okay. If I take the horizontal component of the Normal force and the horizontal component of the applied force and add them together, will that give me a total horizontal force. (498.299cos23)=458.6866 N (1380sin23)=539.208 N So, 997.8946 N?