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3 years ago

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A perfectly conductive plate is placed in the y z-plane. An electromagnetic wave with electric field is incident on the conducto

r. If the wave strikes the plate at t = 0, what are the directions of the electric and magnetic fields of the reflected wave immediately after reflection?

1 answer:

Marizza181 [45]3 years ago

7 0

Answer:

Electric field will move in y-direction, while magnetic field will move in z-direction.

Explanation:

If we make a sketch of electromagnetic wave propagation, we will discover that, the direction of the electric field will be towards y-direction and the direction of the magnetic field will be towards z-direction.

Therefore, if electromagnetic wave with electric field is incident on the conductor, immediately after reflection, Electric field will move in y-direction, while magnetic field will move in z-direction.

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Lexy throws a dart with an initial velocity of 25 m/s at an angle of 60° relative to the ground. what is the approximate vertica

Serjik [45]

An initial velocity is:
v o = 25 m/s
The vertical component of the initial velocity:
v o y = v o * sin 60° =
= v o * √3 / 2 = 25 m/s * √3 / 2 = 21.65 m/s
Answer:
The approximate vertical component of the initial velocity is 21.65 m/s.

7 0

3 years ago

Read 2 more answers

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

3 years ago

Match the theory to the statement that best describes it. 1. big bang. 2. steady state. 3.osscillating universe. 4. inflation Ch

Gre4nikov [31]

Explanation :

(1) Big bang : (1) The most accepted theory on the origin of the universe.

This theory shows the expanding of the universe from high density and high-temperature states.

(2) Steady state : (3) All is the same and will always stay the same.

Steady state means that the properties of any system remain the same always.

(3) Oscillating universe : (4) Agrees with the big bang theory but insists the universe expanded much quicker.

Oscillating universe theory is the result of big bang theory.

(4) Inflation Choices : (2) it's like an inflating and deflating balloon that never stops.

In cosmology, cosmic inflation or deflation is just the expanding and contraction of the universe.

So, the statements and the choices are related as:

(1)-(1)

(2)- (3)

(3)-(4)

(4)-(2)

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3 years ago

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A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.77

Vikki [24]

The formula for the rotational kinetic energy is

$KE_{rot} = \frac{1}{2}(number \ of\ propellers)( I)( omega)^{2}$

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

$I=m L^{2}=(2.67 \ kg) (0.777 \ m)^{2} =2.07459 \ kgm^{2}$

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

$KE_{rot} =( \frac{1}{2} )(5)(2.07459 \ kgm^{2}) (60\ rad/s)^{2}$

$KE_{rot} =18,671.31 \ J$

6 0

3 years ago

Part A Determine the magnitude of the x component of F using scalar notation. Fx F x = nothing lb Request Answer Part B Determin

maksim [4K]

As we know that force F makes an angle of 60 degree with X axis

so the X component is given as

$cos60 = \frac{F_x}{F}$

now we have

$F_x = F cos60$

$F_x = 0.50 F$

Similarly we know that force F makes an angle of 45 degree with Y axis

so the X component is given as

$cos45 = \frac{F_y}{F}$

now we have

$F_y = F cos45$

$F_y = 0.707 F$

Now for the component along z axis we know that

$F_x^2 + F_y^2 + F_z^2 = F^2$

now plug in all components

$(0.707 F)^2 + (0.50 F)^2 + F_z^2 = F^2$

$0.5 F^2 + 0.25 F^2 + F_z^2 = F^2$

$F_z^2 = F^2(1 - 0.75)$

$F_z^2 = 0.25 F^2$

$F_z = 0.5 F$

5 0

3 years ago