Answer:
E = 4.72 * 10⁻⁶ Nm²
Explanation:
Parameters given:
Outer radius, R = 3.70cm = 0.037m
Inner radius, r = 3.15cm = 0.0315m
Permittivity of free space, ε₀ = 8.85 * 10⁻¹² C/Nm²
Charge density: 1.22 * 10⁻³ C/m³
The question requires that we solve using Gauss law which states that the net electric field through a closed surface is proportional to the enclosed electric charge.
Hence,
E = Q/Aε₀
Charge Q is given as
Q = ρπ(R² ⁻ r²)L
A = 2π(R - r)L
E = [ρπ(R² ⁻ r²)L]/[2π(R - r)ε₀L]
Using difference of two squares,
(R² ⁻ r²) = (R + r)(R - r)
E =[ρ(R + r)]/(2ε₀)
E = [1.22 * 10⁻³ *(0.0370 + 0.0315)]/(2 * 8.85 * 10⁻¹²)
E = 4.72 * 10⁻⁶ Nm²
If we are talking on the force being exerted by a segment of a rope of lenght R on the right on a point M which is being also pulled from the Left by a segment of rope R as shown in the figure attached. Then we invoke Newton's Third Law:
"Any force exerted by an object (in this case a segment of the rope) also suffers a equal and opposite force".
If we pick

whis is the tension exerted by the right segment then the left segment will also exert an equal and opposite force so we have that
Answer:
The Momentum Calculator uses the formula p=mv, or momentum (p) is equal to mass (m) times velocity (v). The calculator can use any two of the values to calculate the third.
Explanation:
Answer:
A. When moving towards a high pressure center the pressure values increase in the equipment
B. This area is called high prison since the weight of the atmosphere on top is maximum
Explanation:
A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values 85.5 kPa.
When moving towards a high pressure center the pressure values increase in the equipment
B) This area is called high prison since the weight of the atmosphere on top is maximum
in general they are areas of good weather
Answer:
F_net = 26.512 N
Explanation:
Given:
Q_a = 3.06 * 10^(-4 ) C
Q_b = -5.7 * 10^(-4 ) C
Q_c = 1.08 * 10^(-4 ) C
R_ac = 3 m
R_bc = sqrt (3^2 + 4^2) = 5m
k = 8.99 * 10^9
Coulomb's Law:
F_i = k * Q_i * Q_j / R_ij^2
Compute F_ac and F_bc :
F_ac = k * Q_a * Q_c / R^2_ac
F_ac = 8.99 * 10^9* ( 3.06 * 10^(-4 ))* (1.08 * 10^(-4 )) / 3^2
F_ac = 33.01128 N
F_bc = k * Q_b * Q_c / R^2_bc
F_bc = 8.99 * 10^9* ( 5.7 * 10^(-4 ))* (1.08 * 10^(-4 )) / 5^2
F_bc = - 22.137 N
Angle a is subtended between F_bc and y axis @ C
cos(a) = 3 / 5
sin (a) = 4 / 5
Compute F_net:
F_net = sqrt (F_x ^2 + F_y ^2)
F_x = sum of forces in x direction:
F_x = F_bc*sin(a) = 22.137*(4/5) = 17.71 N
F_y = sum of forces in y direction:
F_y = - F_bc*cos(a) + F_ac = - 22.137*(3/5) + 33.01128 = 19.72908 N
F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N
Answer: F_net = 26.512 N