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1. find the component of the force downwards from which to calculate the normal reaction. 527* sin(32) add the weight force N= 527*sin(32) + 374. 2. Find the frictional force created. Fr = ( 527*sin(32) + 374) * 0.58 ( ~379 N) 3. find the component of the force along the floor to get the...

2 Answers · Science & Mathematics · 28/02/2018

... the string back? W = x*F/2 = 60*0.6 = 36 joule W =

**k**/2*x^2 = F/2x*x^2 = F/2*x = 60*0.6 = 36 joule2 Answers · Science & Mathematics · 27/02/2018

distance a to center of ring is 1.8 [metres?] distance b to center is 3.4 [metres?] distance ebtween a and b is 1.4 [metres?] One of those is wrong, don't you think? I'm assuming: distance a to center of ring is 1.8 m distance b to center...

1 Answers · Science & Mathematics · 24/02/2018

... 0.1m Add the two field values as vectors. E1 =

**k**(6µ)/0.1² E2 =**k**(3µ)/0.1² they are...1 Answers · Science & Mathematics · 20/02/2018

...the two protons at the nucleus = 2*1.6*10^(-19) C E = =

**k***q/ (r^2) = (9*10^9) *3.2*10^(-19) / (26.5*10^(-12))^2 = 4.1*10^12 N/...1 Answers · Science & Mathematics · 20/02/2018

...9.8 m/s²/g a = 2.9509... a = 3.0 g's c) What spring constant,

**k**, for full dunk 60 m below? ½kx² = mgh ½k(60 - 15...2 Answers · Science & Mathematics · 11/02/2018

V = V1 + V2 + V3 V1 = V2 =

**k**∫dq / r =**k**∫λ dr / r =**k**λ ln3 V3 =**k**∫dq / R =**k**∫λ dr...1 Answers · Science & Mathematics · 08/02/2018

...null then must be q + qa = 0 q = - qa = - 7.5 nC b) if r > b E =

**k**(q + qa + qb) / r^2 =**k**qb / r^2 E = - 9x10^9*24.5x10^-9 / (18x10^-2)^2 = - 6...1 Answers · Science & Mathematics · 07/02/2018

Precisely the same answer as yesterday: The y-component of that field, which is 1.36 x 10^3 N/C, multiplied by the area of the circle, which is pi*r^2.

2 Answers · Science & Mathematics · 05/02/2018

a)

**k**was not affected but the measured T were ... mass, and spring constant. T² = 4π²(M+m) /**k**where m is the mass of the spring.2 Answers · Science & Mathematics · 30/01/2018

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