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Stels [109]
3 years ago
15

The electric potential V in the space between two flat parallel plates 1 and 2 is given (in volts) by V = 2200x2, where x (in me

ters) is the perpendicular distance from plate 1. At x = 1.7 cm, (a) what is the magnitude of the electric field and (b) is the field directed toward or away from plate 1?
Physics
1 answer:
Bezzdna [24]3 years ago
6 0

Answer:

a) 74.8 V/m

b) direction -(i)

Explanation:

For this case, the electric field is defined as:

E=(\frac{dV}{dx})

because V is a function dependent of x so we can derivate it:

\frac{dV}{dx}=\frac{d(2200x^2)}{dx}=2*2200*x

so:

E=-4400x\\E=-4400V/m^2*0.017m=-74.8V/m

Because of the sign, the magnitude of electric field 74.8 V/m on the negative direction of i vector, thus means towards the plate 1.

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Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E
Kay [80]

Answer:

The true course: 40.29^\circ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

  • V_w = velocity of wind = 30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h
  • V_p = velocity of plane in still air = 100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h

Assume:

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Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h

Let us find the direction of this resultant velocity with respect to east direction:

\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ

This means the the true course of the plane is in the direction of 40.29^\circ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h

Hence, the ground speed of the plane is 96.68 km/h.

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