Recent content by tomrule007

so are u saying that is right?

after looking it up i found power = force x velocity (is this what your talking about ? it was never mentioned in my class)

Homework Statement Ignoring air resistance is a large error when dealing with cars, especially when they are traveling at high speeds. Remember that air resistance depends on the speed squared F_ar=-Av^2. A) draw a "FBD" for the car. show that the power required to keep a car going at a...

Just curious if anyone could tell me if i have the right idea -Thanks Tom

Homework Statement A bullet of mass m moving with velocity V_o crashes into a block of mass M at rest on a frictionless horizontal surface the block is sitting at the edge of a table that is H high. When the bullet hits the block it flies off the table and it lands a distance x from the bottom...

Thanks for all your help

so for part D i can just use the difference in Kinetic energy? (1/2)(.2)(15)^2 - (1/2)(.2)(12)^2 = Energy lost due to heat (.1)(225) - (.1)(144)= 8.1 N Well that seems simple if i did it right. And i just want to make sure that on part B when it says "find the components of the impulsive...

just updated it with the time now i got to work on C)what is the direction of the impulsive force with respect to the barrier? I can solve for this by taking the arctan of the two impulse components ? Tan x = opposite / adjacent = 81.8591/ -2.1678 x=-88.483 so this means almost all the...

So this is the same thing i did before but updated with the new signs B) I_x=.2(8.48528-9.02723)= -.10839 I_y=.2(8.48528-(-11.9795))= 4.092996and to solve for the F_average i just do divide by .05 so F_x=-.10839/.05 = -2.1678 N F_y=4.092996/.05 = 81.8591 N

Ok now let's see if i get it A)Before: V_x=15cos53 = 9.02723 (going to the right) V_y=15sin53 = -11.9795 (going down) After: V_x=12cos45 = 8.48528 (going to the right) V_y=12sin45 = 8.48528 (going up) For part B I'm still a little confused because if F_average = M(change in v/change in t)...

Homework Statement A 200g puck sliding on ice strikes a barrier at an angle of 53degree and bounces off at an angle of 45degree. The speed of the puck before the bounce was 15m/s and after the bounce it is 12m/s. The time of contact during the bounce is .05seconds A) write these velocities in...

I was thinking about that and thought i had it right but i just relized i think it is coming back from the ramp. 0 -> 1 = from spring to ramp 1 -> 2 = from ramp to spring 2 -> 3 = from spring to ramp so going to an odd number = going towards ramp and going to even number = going toward...

o and when i check the friction using the distance 5.7398 the force comes out to equal 2.25 which makes me think it is right since i want the energy to = 0

Thanks for all the help now i just got few quick questions for the parts b,c,d (I think i got the answers already)B) The mass comes down and goes over the rough patch again on the way back to the spring. how many passes will the mass make over the patch before running out of energy? My answer...

o that's a mistake so if i use .2kg i get E_friction= (.5)(.2)(.2)(9.8) = .196 Then PE_gravity 2.25-.196 = mgy Y=2.054/mg = 1.04796 y= 1.04796 (so that should be the anwser if i did everything right)