number 2 question What's MoreActivity 3. 4. LET'S COMPUTE...On a separate sheet, illustrate the free-body diagram and calculate what isasked in each situation.1. A car that weighs 2.0 x 104 N is resting on the ramp of a tow truck. The car'sbreak transmission lock is released and only the chain prevents the car from rollingbackward off the tow truck. The ramp makes an angle of 30° with the horizontal.Find the tension in the chain and the force with which the ramp pushes on thecar's tires.2. An engine with weight w hangs from a chain that is linked at a point O to twoother chains, one is connectedto the ceiling and the otherone to the wall. Find thetension in each of the threechains, assuming that w isgiven and the weights of thechains are negligible. Theangle of separation betweenthe wall and the ceiling is 45°.We replace the weightby its components.sin aw cos e(a) Car on ramp(b) Free-boby diagram for car

Draw a free-body diagram of the system.

2. An engine with weight w hangs from a chain that is linked at a point O to two other chains, one is connected to the ceiling and the other one to the wall. Find the tension in each of the three We replace the weight by its components. chains, assuming that w is given and the weights of the chains are negligible. The angle of separation between the wall and the ceiling is 45°. sin e w cos e (a) Car on ramp (b) Free-boby diagram for car

2. An engine with weight w hangs from a chain that is linked at a point O to two other chains, one is connected to the ceiling and the other one to the wall. Find the tension in each of the three We replace the weight by its components. chains, assuming that w is given and the weights of the chains are negligible. The angle of separation between the wall and the ceiling is 45°. sin e w cos e (a) Car on ramp (b) Free-boby diagram for car

Inquiry into Physics

8th Edition

ISBN:9781337515863

Author:Ostdiek

Publisher:Ostdiek

Chapter4: Physics Of Matter

Section: Chapter Questions

Problem 17Q: (Indicates a review question, which means it requires only a basic understanding of the material to...

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Oh no! Our expert couldn't answer your question. Don't worry! We won't leave you hanging. Plus, we're giving you back one question for the inconvenience. Here's what the expert had to say: Hi and thanks for your question! This is a writing assignment that is out of scope of Q&A. If you need writing help, please click Contact Research (Essays) link at the bottom of the page. We've credited a question back to your account. Apologies for the inconvenience. Ask Your Question Again 7 of 10 questions left until 1/22/21 Question Hello,I wanna asked if my answer is correct... For A to B I am using Ta+Va=Tb+Vb And I get Ta= 0 since there is no velocity Tb = ½(50)(VB)²=25(VB)² Va=½ks² = ½(2000)(0.6)² =600N Vb=mgh =(50)(9.81)(0.9)= 441.45 Then Vb =2.52 m/s And then using vertical motion from b to c to find t Y=Y initial+(V sin36.87)tBC+1/2a(tBC)² Then t = 0.609s And find d using d=(vcos36.87)t d=1.23m Is my answer and method correct? Sorry for my bad english