Answer:
a = 4.9(1 - sinθ - 0.4cosθ)
Explanation:
Really not possible without a complete setup.
I will ASSUME that this an Atwood machine with two masses (m) connected by an ideal rope passing over an ideal pulley. One mass hangs freely and the other is on a slope of angle θ to the horizontal with coefficient of friction μ. Gravity is g
F = ma
mg - mgsinθ - μmgcosθ = (m + m)a
mg(1 - sinθ - μcosθ) = 2ma
½g(1 - sinθ - μcosθ) = a
maximum acceleration is about 2.94 m/s² when θ = 0
acceleration will be zero when θ is greater than about 46.4°
Answer:
E = 1.655 x 10⁷ N/C towards the filament
Explanation:
Electric field due to a line charge is given by the expression
E = ![[tex]\frac{\lambda}{2\pi\times\epsilon_0\times r}](https://tex.z-dn.net/?f=%5Btex%5D%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Ctimes%5Cepsilon_0%5Ctimes%20r%7D)
[/tex]
where λ is linear charge density of line charge , r is distance of given point from line charge and ε₀ is a constant called permittivity and whose value is
8.85 x 10⁻¹².
Putting the given values in the equation given above
E = 
E = 1.655 x 10⁷ N/C
Answer:
42.6 m
Explanation:
mass of crate m = 53 kg
coefficient of kinetic friction, μ = 0.36
acceleration due to gravity, g = 9.8 m/s^2
Force, F = 372.098 N
Net force, f = F - friction force
f = 372.098 - μ m x g = 372.098 - 0.36 x 53 x 9.8
f = 185.114 N
acceleration, a = f / m = 185.114 / 53 = 3.49 m/s^2
initial velocity, u = 0
time, t = 4.94 s
s = ut + 1/2 at^2
s = 0 + 1/2 x 3.49 x 4.94 x 4.94
s = 42.6 m
