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
D. 5.098 x 106 is the correct answer when reduced to the proper notation.
<span>high pressure produced by the clouds because its the most likely!!!!!!!!!!</span>
Answer:
650.65 K or 377.5°C
Explanation:
Area = A = 10 m²
Thickness of wall = L = 2.5 cm = 2.5×10⁻² m
Inner surface temperature of wall = 
= 415°C = 688.15 K
Outer surface temperature of wall = 
Heat loss through the wall = 3 kW = 3×10³ W
Thermal conductivity of wall = k = 0.2 W/m K
Assumptions made here as follows
- There is not heat generation in the wall itself
- The heat conduction is one dimensional
- Heat flow follows steady state
- The material has same properties in all directions i.e., it is homogeneous.
Considering the above assumptions we use the following formula
∴ The temperature of the outer surface of the wall is 650.65 K or 377.5°C
Answer:
2.5 m/s
Explanation:
The velocity of the package relative to the ground = the velocity of the package relative to the helicopter + the velocity of the helicopter relative to the ground
v = 0 m/s + 2.5 m/s
v = 2.5 m/s
At the moment it is released, the package is rising at 2.5 m/s.
