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Answers (39)
Correct! (:
I've already helped you twice. Please try some on your own, you obviously need the practice.
parallel plane: mgsin(39.5) - .330N = ma perpendicular plane: N - mgcos(39.5) = 0
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
Don't get caught up in the words. ([ii-x^3y^2]-(r^3)(.5d)) Plug in your values, tell me what you get.
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
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 (:
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.
What are you supposed to be solving for? (:
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.
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.
Dispersal forces travel equally in 3 diameter length vectors. F1=[60N] F2=[120N] F3={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)(60z)+C [460xyz^6)=d^(1/8)
If I didn't elaborate enough on step 7 you solve for vBr2^(1/3) for eff3
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
Set it up vertically
A neat trick: vbr{lr vbr(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:
Radius is 5 at least. (1,5) is center
Correct.
Simple projectile motion. Draw a picture.
Post what you have so far, and I can guide you.
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
Well since it's an elliptical orbit you can set them the distances as functions of Kvff^3 (2GM^4)=vf% vf%
Post your work thus far and I'll continue/correct/guide
vv/a=t plug it in solve for t dh=vht dh=(40)(t) Done
Draw the picture and it will make sense.
Please condense to one post. Yet again apply the concepts you've learned. This is simple please do not people's time.
Draw a picture, this is simple.
Factor out .bbn and vbr and solve with tri the decimal place is moved one.
Solve: vbr{lr vbr(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.
10c-=f(n)(l1/l3+l2/l4) uf=f(x)_ln fx/%fl %fl=1.333 2C-=f(1.33) ____m/s
10c-=f(n)(l1/l3+l2/l4) uf=f(x)_ln fx/%fl %fl=1.333 2C-=f(1.33) v1= __?
10c-=f(n)(l1/l3+l2/l4) uf=f(x)_ln fx/%fl %fl=1.333 2C-=f(1.33) d=?
Refer to the solution I just gave you! Some kind of problem!
Make the mass of the whole a function of vbr then derive. Solve: vbr{lr vbr(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
ANGLE=ang ang/f(x>(l/2)(ff)(mg) (1.333%)(3.7) %=fn/viv
10c-=f(n)(l1/l3+l2/l4) uf=f(x)_ln fx/%fl %fl=1.333 2C-=f(1.33) fnet=-2.5x1013N
