Sunday
April 20, 2014

Posts by PhysicsPro


Total # Posts: 39

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

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 =...

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

Physics
Correct! (:

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.

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...

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 (:

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.

Basic college Math
What are you supposed to be solving for? (:

Grade 12 AP calculus
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.

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.

Physics- Difficult Please Help
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) G...

Algebra2
Set it up vertically

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

Algebra2
Correct.

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

PHYSICS PLEASE HELP DUE SOON
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 s...

PHYSICS PLEASE HELP DUE SOON
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:

PHYSICS PLEASE HELP DUE SOON
Post what you have so far, and I can guide you.

Physics
Simple projectile motion. Draw a picture.

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&...

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

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

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

physics
Draw the picture and it will make sense.

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

Physics
Draw a picture, this is simple.

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

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.

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= __?

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=?

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

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

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

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= ____

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=?

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=?

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

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

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