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Arte-miy333 [17]
3 years ago
15

An electromagnetic wave is propagating in the positive x direction. at a given moment in time, the magnetic field at the origin

points in the positive y direction. in what direction does the electric field at the origin point at that same moment?
Physics
2 answers:
antiseptic1488 [7]3 years ago
6 0

Since in an electromagnetic wave the electric and magnetic fields are perpendicular to each other and perpendicular to the direction of motion, the electric field has to point in the z direction.

Alika [10]3 years ago
5 0

Answer:

Along negative z axis

Explanation:

As we know that in any electromagnetic waves propagation the electric field component and magnetic field component is perpendicular to each other and the propagation of wave is perpendicular to both the components of waves.

so here it follows the condition that velocity is perpendicular to both electric field and magnetic field so we will have

\vec v = \vec E \times \vec B

since velocity is towards positive x direction and magnetic field is towards positive y direction so for the above condition to be hold good the electric field must be along negative z direction.

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A projectile is fired with an initial speed of 65.2 m/s at an angle of 34.5º above the horizontal on a long flat firing range. (
Margarita [4]

Answer:

A) h = 69.58 m

D) v = 58.12 \frac{m}{s}  (Speed magnitude)

α = 22.49° (Speed direction above the horizontal)

Explanation:

Conceptual analysis:

To solve this problem we consider the following concepts:

1) The projectile in its movement describes a curved line called a parabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane.

The initial velocity (Vo) is tangent to the trajectory at the initial point and can be broken down into two components, one vertical (Voy) and one horizontal(Vox):

V_{ox} =  V_{o}Cos\alpha _{o} Formula (1)

V_{oy} =  V_{o}Sin\alpha _{o} Formula (2)

Where:

Vo: Initial velocity in m/s

\alpha_{o}: Initial angle above the horizontal in grades

2) The formula to calculate its velocity at any vertical position(y) is as follows:

v_{y}^{2} = v_{oy}^{2} -2gy Formula (3)

Where:

v_{f}^{2}: Final speed component in vertical direction in m/s

v_{oy}^{2}: Initial speed component in vertical direction in m/s

g: acceleration due to gravity in m/s2

3) The formulas to calculate the projectile velocity components at any time (t) are:

v_{x} = v_{ox} Because the movement is uniform in the x direction (constant speed)

v_{y} = v_{oy}-g*t Formula (4)  Because the movement is uniformly accelerated in the y direction

Known information:

We know the following data:

v_{o} = 65.2 \frac{m}{s}

\alpha _{o} = 34.5º above the horizontal

g=9.8 \frac{m}{s^{2}}

Development of the problem:

Initial speed components(Vox, Voy), (Formula (1), Formula (2)

v_{ox} =  65.2*Cos34.5 = 53.7\frac{m}{s}

v_{oy} =  65.2*Sin34.5 = 36.93\frac{m}{s}

A) Maximum height (h):

When the projectile reaches its maximum height (h) ,the speed component Vy = 0, then, we replace this value in the Formula (3):

0=(36.93)^2-2*9.8*h

h=\frac{ (36.93)^2}{2*9.8} = 69.58 m

D) Speed of the projectile 1.50s after firing

We replace t=1.5 s in the formula(4)

v_{y} = 36.93-9.8*1.5 = 22.23 \frac{m}{s}

v_{x} = v_{ox} = 53.7 \frac{m}{s}

v = \sqrt{53.7^{2}+22.23^{2}} = 58.12 \frac{m}{s}  (Speed magnitude)

\alpha = tan^{-1} (\frac{v_{y}}{v_{x}}) =  tan^{-1} (\frac{22.23}{53.7})

α = 22.49° (Speed direction above the horizontal)

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Answer:

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Explanation:

patron

Explanation:

 The patron is something or someone who defends some cause or point of view. In the art field, for example, the patron may be considered a sponsor, that is, someone who is known for defending a particular group of people or specific situation.

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7 0
1 year ago
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