Recent content by ideefixem

Ahhh yes! Applied force - F2 = 1 kg (2m/s^2) 12N-10N=2N I believe that I finally understand what you were trying to explain. Thank you so much for your help and patience... It's been a long weekend.

Errr... This is painful... Thank you to the repliers, especially saket. You told me that I did arrive to the correct force between 2-3, 6 N. Seeing as the net force equals 12 like you said, would that have to lead me to believe that 12-6 equals Fnet-F2= F1= 6? Thank you for being so patient.

Would the force exerted between 2-3 = f=ma x=3(2) x=6 N Force between 1-2 x= 2(2) x= 4 N Seems incredibly simple so I'm sure there must be some sort of problem somewhere along the way.

Well finding the acceleration of the system was easy enough/ f=ma 12=6(x) x=2 I drew the fbd however I'm having trouble differentiating between the different blocks. For instance, Block 1- 1 kg acting on block 2- 2 kg... When I use the formula f=ma clearly I will arrive at two different...

Homework Statement Blocks with masses of 1 kg, 2 kg, and 3 kg are lined up in a row on a frictionless table. All three are pushed forward by a 12 N force applied to the 1 kg block. How much force does the 2 kg block exert on the 3 kg block? How much force does the 2 kg block exert on the...

I'm using the distance from the Earth to the Sun. 1.50 x 10^11 = r r^2 = 2.25*10^22

How about this... ((6.67*10^(-11)) * (1.99*10^30))/ (2.25*10^22) = 0.005899244 m/s^2 ~0.00

Homework Statement What is the acceleration due to gravity of the sun at the distance of the Earth's orbit? Homework Equations law of gravity = Gm/r^2 Sun's Mass: 1.99x10^30 kg, earth-sun distance: 150x10^6 km The Attempt at a Solution ((6.67*10^(-11)) * (1.99*10^30))/...

Hmm... That wasn't even close... Let me try this again. cos30(7)=vx= 6.06 m/s sin30(7)=vy= 3.5 m/s -1=3.5(t)+.5(-9.8)(t^2) 0=-4.9t^2+3.5t+1 t= 0.933 s x= (6.06)(.933)+.5(0)(.933)^2 x= 5.65 m How does that look?

Thank you very much for your help... I think this is correct... could you please review it for me? cos40(.8)= ax= 0.613 m/s^2 sin40(.8)= ay= 0.514 m/s^2 Vfx= 5+(0.613)(6)= 8.68 m/s Vfy= 5+(0.514)(6)= 8.08 m/s V= sqrt(8.68^2+8.08^2)= 11.86 m/s xf=0+5(6)+.5(.613)(6)^2= 41.034 m...

Homework Statement A skateboarder starts up a 1.0 m high 30 degree ramp at a speed of 7.0 m/s. The skateboard wheels roll without friction. How far from the end of the ramp does the skateboarder touch down? The Attempt at a Solution ramp length= h=1sin30=1/2 length=2m v^2=vi^2+2ad...

Homework Statement A sailboat is traveling east at 5.0 m/s. A sudden gust of wind gives the boat an acceleration of 0.80 m/s^2 at 40 degrees north of east. What are the boat's speed and direction 6.0 s later when the gust subsides? Homework Equations xf=xi+vix(delta t)+1/2ax(delta t)^2...

Haha, quite the memory. Yep, taken right from the book.

Homework Statement A stubborn 120 kg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around the mule and pulls with his maximum force of 800 N. The coefficients of friction between the mule and the ground are mu,s = 0.8 and mu,k = 0.5. Is...

Thank you for your help.