The ball would move about, 674 kilometers. I think I’m correct. Somebody correct me if I’m not.
Answer:
E = 9.4 10⁶ N / C
, The field goes from the inner cylinder to the outside
Explanation:
The best way to work this problem is with Gauss's law
Ф = E. dA = qint / ε₀
We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.
The flow on the faces is zero, since the field goes in the radial direction of the cylinders.
The area of the cylinder is the length of the circle along the length of the cable
dA = 2π dr L
A = 2π r L
They indicate that the distance at which we must calculate the field is
r = 5 R₁
r = 5 1.3
r = 6.5 mm
The radius of the outer shell is
r₂ = 10 R₁
r₂ = 10 1.3
r₂ = 13 mm
r₂ > r
When comparing these two values we see that the field must be calculated between the two housings.
Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is
λ = q / L
Qint = λ L
Let's replace
E 2π r L = λ L /ε₀
E = 1 / 2piε₀ λ / r
Let's calculate
E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3
E = 9.4 10⁶ N / C
The field goes from the inner cylinder to the outside
Answer:
you know what I think you should not give these kind of q
Answer:
a) F = (137.4 i ^ + 185 j ^) N
b) F = 230.2 N , θ = 53.5º
Explanation:
In this exercise we ask to find the net force, for which we will define a coordinate system fix the donkey and use trigonometry to decompose the forces
Jack F₁ₓ = 63.9 N
Jill F₂ = 79.1 N with direction 45º to the left
cos (180 -45) = F₂ₓ / F₂
sin 135 =
/ F₂
F₂ₓ = F₂ cos 135
F_{2y} = F₂ sin 135
F₂ₓ = 79.1 cos 135 = -55.9 N
F_{2y} = 79.1 sin 135 = 55.9 N
Jane F₃ = 183 N direction 45th to the right
cos 45 = F₃ₓ / F3
sin 45 = F_{3y} / F3
F₃ₓ = F₃ cos 45 = 183 cos 45
F_{₃y} = F₃ sin 45 = 183 sin 45
F₃ₓ = 129.4 N
F_{3y} = 129.4 N
we add each component of the force
Fₓ = F₁ₓ + F₂ₓ + F₃ₓ
Fₓ = 63.9 + (-55.9) + 129.4
Fₓ = 137.4 N
F_{y} = F_{2y} + F_{3y}
F_{2y} = 55.9 + 129.4
F_{2y} = 185.3 N
we can give the result of the forms
a) F = (137.4 i ^ + 185 j ^) N
b) in the form of module and angle
F = RA (Fₓ² + F_{y}²)
F = Ra (137² + 185²)
F = 230.2 N
tan θ = F_{y} / Fₓ
θ = tan⁻¹ F_{y} / Fₓ
θ = tan⁻¹ (185/137)
θ = 53.5º
Answer: actually to solve this u must know what 3.2m/s2 is..do u know it ??...wel let me tell...3.2m/s2 means that in every second the speed of the body increases by 3.2m/s..so if it acts for 6.8 seconds..increase in speed will be 3.2*6.8 however note that this is the increase in speed.. have to change this in km/hr & add this to the initial speed(27km/hr) to get the answer....hope this has cleared ur doubts....
Explanation: