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**k**= spring constant = 80 N/m m = mass = 0.5 kg L = distance from A to B = 2.5 m R = radius of...m)) E = 20.58 J That is the initial energy that must be stored in the spring. E =**k**×d²/2 (20.58 J) = (80 N/m)×d²/2 d = √[2×(20.58 J)/(80 N/m)] d = 0.7173 m1 Answers · Science & Mathematics · 25/01/2013

**K**= (1/2) I omega^2 + (1/2) m v^2 , and omega = v/r I = (2/3) mr^2 => m = (3/2) I...20.32 m/s (1/2) mv^2 = (1/2)(0.276 kg)(20.32 m/s)^2 = about 50 J but use calculator. Looks like the**K**may be equally split among rotational and translational energies. (b) 20.32 m/s, from above (c) 95 J - (mg...1 Answers · Science & Mathematics · 27/11/2012

### 10 points for any help on this one!!! tried this pne out for sometime...keep getting wrong answer!!?

The spring constant

**k**= 720/(0.55*10^-3) = 1309*10^3 N/m Your energy at 1.3 m M... compressed, and the energy in the spring is E = 0.5***k***x^2, where x = compression 0.5***k***x^2 = 957.6 (x in meters) x = √[2*957...2 Answers · Science & Mathematics · 18/03/2008

Step by step answers posted here http://www.maths

**k**ey.com/question2answer/1447/compute-%E2%88%87-x-a-for-a-vector as**k**your physics homewor**k**questions at http://www.maths**k**ey.com/question2answer/… and get free math help. all the best1 Answers · Science & Mathematics · 19/02/2013

1.8

**k**(omega) = 1,800 omega 1M omega = 1,000,000 omega1 Answers · Science & Mathematics · 16/06/2012

Use the equation E = hc/lambda you will need to convert from eV to J (1eV = 1.6E-19J) The minimum voltage to be applied is 8.968 kV.

2 Answers · Science & Mathematics · 13/12/2012

wht is s , n and d???

1 Answers · Science & Mathematics · 31/03/2013

**k**= F / x = 16lb / (8/9)ft = 18 lb·ft ω = √(**k**/m) = √(18lb·ft / (16lb / 32ft/s²)) = 6 rad/s So we can model...1 Answers · Science & Mathematics · 25/06/2016

**k**= spring constant ( kg/sec^2) x= compression( extension) meters F=force in kg -m/sec^2 F= kx PE1 = 0.5kx^2 = 0.5(**k**) [0.042^2] kg-m = stored energy before spring is released...3 Answers · Science & Mathematics · 20/02/2012

Electric field =

**k*** q / r^2 for each particle (where**k**is a constant, 1 / (4 * pi * permittivity of...0.2-r) away from the other one. Now add up the electric fields:**k*** q1 / r^2 -**k*** q2 / (0.2-r)^2 = 0 (it's minus because they point in opposite...1 Answers · Science & Mathematics · 26/05/2009