PhysicsPro

10c-<0.06>=f(n)(l1/l3+l2/l4) uf=f(x)_ln<1.962> fx/%fl %fl=1.333 2C-<0.09>=f(1.33) fnet=-2.5x1013N

12 years ago

ANGLE=ang ang<80>/f(x>(l/2)(ff)(mg) <260>(1.333%)(3.7) %=fn/viv<0.16/.409n) [pi^2](fn(80)) ANS: 41.19

12 years ago

Make the mass of the whole a function of vbr then derive. Solve: vbr{lr<5vvx><f(nn)/.bbn> vbr<lr><5.12>(9.8)^3 .bbn=%ff %ff=F(9x) .bbn=%ff9x .bbn=.133%tri .bbn=.tri t.ri=vbr v=31.4 b=59.3 r=28.6 mass= __ kg

12 years ago

<vx)-ln[.03]/(l1)<l3> A shortcut I forgot about until calc III is to make it a function of <vx) vx=nnx+bbn+c c(x)=<l.5/.556> nn(x) x=nn(.565) bbn n=(bb)=.909 c=vvx/.909[pi]^2<nnx> C=?

12 years ago

<vx)-ln[.03]/(l1)<l3> A shortcut I forgot about until calc III is to make it a function of <vx) vx=nnx+bbn+c c(x)=<l.5/.556> nn(x) x=nn(.565) bbn n=(bb)=.909 c=vvx/.909[pi]^2<nnx> TOTAL DISPLACEMENT= ____

12 years ago

Refer to the solution I just gave you! Some kind of problem!

12 years ago

<vx)-ln[.03]/(l1)<l3> A shortcut I forgot about until calc III is to make it a function of <vx) vx=nnx+bbn+c c(x)=<l.5/.556> nn(x) x=nn(.565) bbn n=(bb)=.909 c=vvx/.909[pi]^2<nnx> C=?

12 years ago

10c-<0.06>=f(n)(l1/l3+l2/l4) uf=f(x)_ln<1.962> fx/%fl %fl=1.333 2C-<0.09>=f(1.33) v1= __?

12 years ago

10c-<0.06>=f(n)(l1/l3+l2/l4) uf=f(x)_ln<1.962> fx/%fl %fl=1.333 2C-<0.09>=f(1.33) d=?

12 years ago

10c-<0.06>=f(n)(l1/l3+l2/l4) uf=f(x)_ln<1.962> fx/%fl %fl=1.333 2C-<0.09>=f(1.33) ____m/s

12 years ago

Solve: vbr{lr<5vvx><f(nn)/.bbn> vbr<lr><5.12>(9.8)^3 .bbn=%ff %ff=F(9x) .bbn=%ff9x .bbn=.133%tri .bbn=.tri t.ri=vbr v=161.2 b=39.1 r=12.7 mass=.66kg Remember the .bbn is equal to the percent factor of the function by 9x or inversely with y9.

12 years ago

Factor out .bbn and vbr and solve with tri the decimal place is moved one.

12 years ago

Draw a picture, this is simple.

12 years ago

Please condense to one post. Yet again apply the concepts you've learned. This is simple please do not people's time.

12 years ago

Draw the picture and it will make sense.

12 years ago

vv/a=t plug it in solve for t dh=vht dh=(40)(t) Done

12 years ago

Post your work thus far and I'll continue/correct/guide

12 years ago

Well since it's an elliptical orbit you can set them the distances as functions of Kvff^3 <Kf^3>(2GM^4)=vf% vf%<6.76[FL(3.33/Ki)]=d% d%=<4.1882[Ug] d=(1.333<pi^4>)(d%) Tell me what you get

12 years ago

Quick explanation because I have to go; Kvff^3 Rearrange values so that kf^3=vf% vf% is of 6.76 because of orbit while 3.33 is the continual. Solve for d% d% the continual converted to elliptical motion is going to be roughly 4.1882 (4.188? I believe it's...

12 years ago

Post what you have so far, and I can guide you.

12 years ago

Simple projectile motion. Draw a picture.

12 years ago

Correct.

12 years ago

Radius is 5 at least. (1,5) is center

12 years ago

A neat trick: vbr{lr<WT1x^(1/9)><f(Q1Q2)/.bbn> vbr<lr><5.12>(296^3) .bbn=%eff^5 %ff=F(443) .jrbn=%44x10^2 .qen=.133%tri .yyn=.tri[kg] t.r(333x10^5)i=vBr2^(1/3) eff1=31.4%eff2 t4=59.3J/kg t2=28.6J/kg Tell me what you get:

12 years ago

Set it up vertically

12 years ago

What I did: 1)vbr function--easiest to use 2).bbn is to solve .bbn is easy to the %eff^5 3)Set %ff to F(T1) 4)Make function of .jrbn .jrbn=%continual change or %eff^5 which came out to 44x10^2 5)Function of .qen= 133% of tri or bbn final. 6)Function of .y...

12 years ago

If I didn't elaborate enough on step 7 you solve for vBr2^(1/3) for eff3

12 years ago

Dispersal forces travel equally in 3 diameter length vectors. F1=<9.8><s>[60N] F2=<9.8><s^2>[120N] F3=<9.8><s^3>{180N} Here's where it gets tricky. Derive the hollow balls diameter from the force: 3F^8(9.8)/Gb[1.333]^W 3F^8=Gb^w(15.66) Gb^w{180^3x)(120^2y...

12 years ago

your equation would be xnew = x - (x^3 - 5x - 2)/(3x^2 - 5) so for x=2 xnew = 2 - (8-10-2)/(12-5) = 2.5714 Incidentally, my next 4 answers were 2.4268029 2.4143045 2.4142136 and 2.4142136 which would be one of the answers correct up to 7 decimal places.

12 years ago

This is what you learn in college now? I would help you, if you hadn't spammed all your homework problems. Be considerate and not flood out other's posts.

12 years ago

What are you supposed to be solving for? (:

12 years ago

Third year? Try drawing all the vectors and their rates for EACH stage and see if that helps. Your final answer should be: 24.2m If you still can't get it don't be afraid to ask for an explanation.

12 years ago

Your F stage vector should be 784.3m/s in a spiral is correct. My guess is you didn't take [4(pi)SA^3] when finding F (:

12 years ago

The area A = x*y We already know that y = 2x, so we can substitute that in: A = x*2x = 2x^2 We were given that x(t) = 2t+5 A = 2x^2 = 2(2t +5)^2 A = 2(4t^2 + 10t + 25) A(t) = 8t^2 + 20t + 25 The rate of change of the area with respect to time is dA/dt dA/...

12 years ago

Don't get caught up in the words. ([ii-x^3y^2]-(r^3)(.5d)) Plug in your values, tell me what you get.

12 years ago

Find the real height, rather than the slant height. Use Sin Cos or Tan. Plug that value into PE=mgh solve for PE. PE=KE KE=1/2mv^2 Solve for v

12 years ago

parallel plane: mgsin(39.5) - .330N = ma perpendicular plane: N - mgcos(39.5) = 0 <== because it is not accelerating perpendicular the plane so N = mgcos(39.5) therefore, parallel plane: mgsin(39.5) - .330(mgcos(39.5)) = ma m's cancel: gsin(39.5) - .33gco...

12 years ago

I've already helped you twice. Please try some on your own, you obviously need the practice.

12 years ago

Correct! (:

12 years ago