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May 1, 2016

Posts by PhysicsPro

Total # Posts: 39

calc
December 19, 2012

Physics
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) - .33gcos(39.5) = a a...
December 19, 2012

physics
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
December 19, 2012

Physics
Correct! (:
December 19, 2012

Physics
Don't get caught up in the words. ([ii-x^3y^2]-(r^3)(.5d)) Plug in your values, tell me what you get.
December 19, 2012

calc
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/dt = 16t + 20 When t=2...
December 19, 2012

Physics University
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 (:
December 2, 2012

Physics University
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.
December 2, 2012

Basic college Math
What are you supposed to be solving for? (:
December 2, 2012

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.
December 2, 2012

College Math
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.
December 2, 2012

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) ...
November 29, 2012

Algebra2
Set it up vertically
November 29, 2012

Algebra2
Radius is 5 at least. (1,5) is center
November 29, 2012

Algebra2
Correct.
November 29, 2012

If I didn't elaborate enough on step 7 you solve for vBr2^(1/3) for eff3
November 29, 2012

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 .yyn=tri what you just ...
November 29, 2012

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:
November 29, 2012

Post what you have so far, and I can guide you.
November 29, 2012

Physics
Simple projectile motion. Draw a picture.
November 29, 2012

College Physics
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 two, but shouldn...
November 27, 2012

College Physics
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
November 27, 2012

College Physics
Post your work thus far and I'll continue/correct/guide
November 27, 2012

physics
vv/a=t plug it in solve for t dh=vht dh=(40)(t) Done
November 27, 2012

physics
Draw the picture and it will make sense.
November 27, 2012

Physics
Please condense to one post. Yet again apply the concepts you've learned. This is simple please do not people's time.
November 19, 2012

Physics
Draw a picture, this is simple.
November 19, 2012

Physics HARD
Factor out .bbn and vbr and solve with tri the decimal place is moved one.
November 19, 2012

Physics HARD
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.
November 19, 2012

Physics
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= __?
November 19, 2012

Physics
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=?
November 19, 2012

Physics
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
November 19, 2012

Physics
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
November 19, 2012

Physics
Refer to the solution I just gave you! Some kind of problem!
November 19, 2012

Physics
<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= ____
November 19, 2012

Math -Calculus
<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=?
November 19, 2012

Physics
<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=?
November 19, 2012

Physics
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
November 19, 2012

physics university
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
November 19, 2012

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