X=.5(a)t^2 can be used: 2.5m=.5(g)(1), g=5m/s^2.
There are eight planets in the milky way galaxy
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)
For this case, the first thing you should do is define a reference system.
Once the system is defined, we must follow the following steps:
1) Do the sum of forces in a horizontal direction
2) Do the sum of forces in vertical direction
The forces will be balanced if for each direction the net force is equal to zero.
The forces will be unbalanced if for each direction the net force is nonzero.
Answer:
Add the forces in the horizontal and vertical directions separately.