Answer:
3.135 kN/C
Explanation:
The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}
Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m
So, E = qz/{4πε₀[√(z² + R²)]³}
E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}
E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}
E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}
E = 6.839 × 10³ Cm²/[1.297 m]³F
E = 6.839 × 10³ Cm²/2.182 m³F
E = 3.135 × 10³ V/m
E = 3.135 × 10³ N/C
E = 3.135 kN/C
Given:
mass: 100 kg
height: 500 m
1 kJ = 1000 J
gravity = 9.8 m/s²
velocity before impact: v = √2gh ; v = √2 * 9.8 m/s² * 500 m ; v = 98.99494 m/s
KE = 1/2 m v²
KE = 1/2 * 100 kg * (98.99494 m/s)²
KE = 490,000 J
Pls. see attachment.
Answer:
move at constant velocity.
Explanation:
Newton's first law (also known as law of inertia) states that:
"when the net force acting on an object is zero, the object will keep its state of rest or if it is moving, it will continue moving at constant velocity".
In the case of the probe, friction in deep space is negligible, therefore when the engine is shut down, there are no more forces acting on the probe: the net force therefore will be zero, so the probe will move at constant velocity.
Answer:
a) Eₓ = - A y + 2B x
, b) Ey = -Ax –C
, c) Ez = 0
, d) The correct answer is 3
Explanation:
The electric field and the electric power are related
E = - dV / ds
a) Let's find the electric field on the x axis
Eₓ = - dV / dx
dV / dx = A y - B 2x
Eₓ = - A y + 2B x
b) calculate the electric field on the y-axis
Ey = - dV / dy
dV / dy = A x + C
Ey = -Ax –C
c) the electric field on the z axis
dv / dz = 0
Ez = 0
.d) at which point the electric field is zero
Since the electric field is a vector quantity all components must be zero
X axis
0 = = - A y + 2B x
y = 2B / A x
Axis y
0 = -Ax –C
.x = -C / A
We substitute this value in the previous equation
.y = 2B / A (-C / A)
.y = 2 B C / A2
The correct answer is 3
