A battery provides the necessary push to make current flow.
C is your answer im pretty sure
Answer:
(a) 1.11sec
(b) 14.37m/s
(c) 31.78m
Explanation:
U = 18m/s, A = 37°, g = 9.8m/s^2
(a) t = UsinA/g = 18sin37°/9.8 = 18×0.6018/9.8 = 1.11sec
(b) Ux = UcosA = 18cos37° = 18×0.7986 = 14.37m/s
(c) R = U^2sin2A/g = 18^2sin2(37°)/9.8 = 324sin74°/9.8 = 324×0.9613/9.8 = 31.78m
No they don't. Without air resistance, objects fall with constant
acceleration.
On Earth, the acceleration is 9.8 m/s² ... the speed of a falling object
at any time is 9.8 m/s faster than it was 1 second earlier.
Answer:
E = k λ₀ / x₀, the field is in thenegative direction of the x axis (-x)
Explanation:
In this problem the electric field of a line of charge is requested, the expression for the electric field is
E = k ∫ dq / r²
where k is the Coulomb constant that you are worth 9 10⁹ N m²/C², that the charge and r the distance to the point of interest, in this case it is the origin (x = 0)
let's use the definite linear density
λ₀ = dq / dx
dq = λ₀ dx
we replace and integrate
E = k λ₀ ∫ dx / x²
E = k λ₀ ( -1 / x)
we evaluate the integral from the lower limit of load x = x₀ to the upper limit x = ∞
E = - k λ₀ (1 /∞ - 1 / x₀)
E = k λ₀ / x₀
as the field is positive the direction is away from the charges, so it is in the negative direction of the x axis (-x)