The correct choice is D .
Hey all you can do is type the question on the internet and click on the one that has the same title as yours and the answers will be there im taking physics and i do that all the answers will be there
Answer:
V = 8.34m/s
Explanation:
Given that
1/2ke^2 = 1/2mv^2 ......1
Where e = 3.75cm = (3.75/100)m
e = 0.0375m
K = 500 N/m
m = 10g = 10/1000
= 0.01kg
Substitute the values into equation 1
0.5×500×(0.0375)^2 = 0.5×0.01×v^2
250×0.001395 = 0.005v^2
0.348 = 0.005v^2
v^2 = 0.348/0.005
v^2 = 69.6
V = √69.6
V = 8.34m/s
The ball launches at the speed of V = 8.34m/s
Greater than is the answer. Pulling an equal force provides no acceleration and no movement. Pulling less than would still cause the object to move to the right.
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)
