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An isotropic point source emits light at wavelength 500 nm, at the rate of 185 W. A light detector is positioned 380 m from the source. What is the maximum rate ∂B/∂t at which the magnetic component of the light changes with time at the detector's location?

Answers

Answer 1

Answer:

Answer:

$(dB)/(dt) = 3.49 *10^(6) \ \ T/s$

Explanation:

Given that

An isotropic point source emits light at a wavelength $\lambda$ = 500 nm

Power = 185 W

Radius = 380 m

Let's first calculate the The intensity of the wave , which is = $(Power )/(Area)$

= $(Power)/(4 \pi r ^2)$

= $(185 \ W)/( 4 \pi (380)^2)$

= $1.0195*10^(-4) \ W/m^2$

Now;

The amplitude of the magnetic field is calculated afterwards by using poynting vector

i.e

$I = ((c)/(2 \mu_0 ))B_(max^2)$

$B_(max^2) = ((2 \mu_0 I)/( c))$

$B_(max^2) = ((2 *4 \pi *10^(-7)*1.0195*10^(-4))/( 3*10^8))$

$B_(max^2) = 8.5409*10^(-19)$

$B_(max) = \sqrt {8.5409*10^(-19)}$

$B_(max) = 9.242*10^(-10)$

The magnetic field wave equation can now be expressed as;

$B = B_(max) sin (kx - \omega t)$

Taking the differentiation

$(dB)/(dt)= - \omega B_(max) \ cos ( kx - \omega t)$

The maximum value ;

$(dB)/(dt) = \omega B _(max)$

where ;

$\omega = 2 \pi f\n\omega = (2 \pi c)/(\lambda)$

then

$(dB)/(dt) = (2 \pi c)/(\lambda) B _(max)$

$(dB)/(dt) = (2 \pi 3*10^8*9.242*10^(-10))/(500*10^(-9))$

$(dB)/(dt) = 3484751.917$

$(dB)/(dt) = 3.49 *10^(6) \ \ T/s$

Answer 2

Answer:

The maximum rate $(∂B/∂t)$ at which the magnetic component of the light changes with time at the detector's location is approximately $6.8 x 10^9$ Tesla per second (T/s).

To find the maximum rate at which the magnetic component of the light changes with time at the detector's location, you can use the formula for the rate of change of magnetic field due to an electromagnetic wave. The formula is given by:

$∂B/∂t = (2π / λ) * E * c$

Where:

$∂B/∂t$ is the rate of change of the magnetic field.

λ is the wavelength of the light.

E is the electric field strength.

c is the speed of light in a vacuum, approximately $3 x 10^8 m/s.$

You have the wavelength (λ) as 500 nm, which is 500 x 10^-9 meters, and the electric field strength (E) can be calculated using the power (P) and the distance (r) from the source. The power emitted by the source is 185 W, and the distance from the source to the detector is 380 m.

First, calculate the electric field strength (E):

$E = sqrt(P / (2π * r^2))$

$E = sqrt(185 W / (2π * (380 m)^2))$

$E = sqrt(185 W / (2π * 144400 m^2))$

$E ≈ 6.325 x 10^-5 N/C$

Now, you can calculate the rate of change of the magnetic field:

$∂B/∂t = (2π / λ) * E * c$

$∂B/∂t = (2π / (500 x 10^-9 m)) * (6.325 x 10^-5 N/C) * (3 x 10^8 m/s)$

$∂B/∂t ≈ (3.77 x 10^15 Hz) * (6.325 x 10^-5 N/C) * (3 x 10^8 m/s)$

$∂B/∂t ≈ 6.8 x 10^9 T/s$

So, the maximum rate at which the magnetic component of the light changes with time at the detector's location is approximately $6.8 x 10^9$ Tesla per second (T/s).

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PLEASE HELP IT'S DUE IN LIKE 2 MINUTES

A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacity of 3000.0 μF and is charged to a potential difference of 60.0 V. Calculate the amount of energy stored in the capacitor. Tries 0/20 Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 6.53 J. Tries 0/20 If the two plates of the capacitor have their separation increased by a factor of 5 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?

Answers

Answer:

1 = 5.4 J

2 = 0.1979 C

3 = 5

Explanation:

Energy in a capacitor, E is

E = 1/2 * C * V²

E = 1/2 * 3000*10^-6 * 60²

E = 1/2 * 3000*10^-6 * 3600

E = 1/2 * 10.8

E = 5.4 J

E = Q²/2C = 6.53 J

E * 2C = Q²

Q² = 6.53 * 2 * 3000*10^-6

Q² = 13.06 * 3000*10^-6

Q² = 0.03918

Q = √0.03918

Q = 0.1979 C

The Capacitor, C is inversely proportional to the distance of separation, D. Thus, if D is increased by 5 to be 5D, then C would be C/5. And therefore, our energy stored in the capacitor is increased by a factor of 5.

A particle located at the position vector m has a force N acting on it. The torque about the origin is

Answers

Final answer:

The torque about a given origin when a force N is acting on a particle at the position vector m is given by the cross product of the position and force vectors. It's represented by the SI unit Newton-meters, and for multiple particles, the total angular momentum is the vector sum of their individual angular momenta.

Explanation:

The torque about a given origin, when a force N is acting on a particle located at the position vector m, is calculated using the cross product of the position vector and the force vector. This can be written as τ = m x N. The SI unit of torque is Newton-meters (N.m).

As an example, if you apply a force perpendicularly at a distance from a pivot point, you will create a torque relative to that point. Similarly, the torque on a particle is also equal to the moment of inertia about the rotation axis times the angular acceleration.

If we consider multiple particles, the total angular momentum of these particles about the origin is the vector sum of their individual angular momenta. This is calculated by the expression for the angular momentum Ỉ = ŕ x p for each particle, where ŕ is the vector from the origin to the particle and p is the particle's linear momentum.

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Final answer:

The torque on a particle at a position vector m with force N acting on it is calculated by taking the cross-product of the position vector and the force. This principle is the same even in systems with multiple particles. The SI unit of torque is Newton-meters (N·m), which should not be confused with Joules (J).

Explanation:

The torque on a particle located at a position vector m with a force N acting on it is calculated by taking the cross-product of the position vector and the force. In terms of physics, torque (τ) is a measure of the force that can cause an object to rotate about an axis, and it is calculated as the product of the force and the distance from the axis of rotation to the point where force is applied. Hence, the formula for torque is τ = r x F where r is the position vector (or distance from the origin to the point where the force is applied) and F is the force. Remember, this equation gives a vector result with a direction perpendicular to the plane formed by r and F and a magnitude equal to the product of the magnitudes of r and F and the sine of the angle between r and F.

The same principle applies to systems where multiple particles are present. The total angular momentum of the system of particles about a particular point is the vector sum of the individual angular momenta about that point. Torque is the time derivative of angular momentum.

The SI unit for torque is Newton-meters (N·m), which should not be confused with Joules (J), as both have the same base units but represent different physical concepts. In this context, a net force of 40N acting at a distance of 0.800m from the origin would generate a torque of 32 N·m at the origin.

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An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's altitude is 532 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.
(b) Relative to the ground, determine the angle of the velocity vector of the package just before impact. (a) Number Units (b) Number Units

Answers

Answer:

$D = 1162.7 \ m$

$\beta =- 65.55^o$

Explanation:

From the question we are told that

The speed of the airplane is $u = 92.3 \ m/s$

The angle is $\theta = 51.1^o$

The altitude of the plane is $d = 532 \ m$

Generally the y-component of the airplanes velocity is

$u_y = v * sin (\theta )$

=> $u_y = 92.3 * sin ( 51.1 )$

=> $u_y = 71.83 \ m/s$

Generally the displacement traveled by the package in the vertical direction is

$d = (u_y)t + (1)/(2)(-g)t^2$

=> $-532 = 71.83 t + (1)/(2)(-9.8)t^2$

Here the negative sign for the distance show that the direction is along the negative y-axis

=> $4.9t^2 - 71.83t - 532 = 0$

Solving this using quadratic formula we obtain that

$t = 20.06 \ s$

Generally the x-component of the velocity is

$u_x = u * cos (\theta)$

=> $u_x = 92.3 * cos (51.1)$

=> $u_x = 57.96 \ m/s$

Generally the distance travel in the horizontal direction is

$D = u_x * t$

=> $D = 57.96 * 20.06$

=> $D = 1162.7 \ m$

Generally the angle of the velocity vector relative to the ground is mathematically represented as

$\beta = tan ^(-1)[(v_y)/(v_x ) ]$

Here $v_y$ is the final velocity of the package along the vertical axis and this is mathematically represented as

$v_y = u_y - gt$

=> $v_y = 71.83 - 9.8 * 20.06$

=> $v_y = -130.05 \ m/s$

and v_x is the final velocity of the package which is equivalent to the initial velocity $u_x$

$\beta = tan ^(-1)[-130.05}{57.96 } ]$

$\beta =- 65.55^o$

The negative direction show that it is moving towards the south east direction

Which equation represents inverse proportionality?Responses

y=m/x
y equals m divided by x

y=mx2+b for nonzero b
y equals m x squared plus b for nonzero b

y=mx
y equals m x

y=mx2

Answers

${ \qquad\qquad\huge\underline{{\sf Answer}}}$

Inverse proportionality means that the two terms should be proportional to inverse of each other ~

That is :

$\qquad \sf \dashrightarrow \: y \propto (1)/(x)$

And according to given options, only option 1 shows inverse proportionality. where y is inversely proportional to x and m is proportionality constant.

$\qquad \sf \dashrightarrow \: y = (m)/(x)$

Final answer:

The equation that represents inverse proportionality in mathematics is y=m/x, where y is inversely proportional to x, and m is the constant of proportionality.

Explanation:

In mathematics, inverse proportionality, or inversely proportional, is a concept where one variable increases when the other variable decreases, and vice versa. It lies at the heart of various mathematical and real-world applications. The equation y=m/x represents inverse proportionality, where 'y' is inversely proportional to 'x'. In this equation, 'm' is the constant of proportionality. As 'x' increases, 'y' decreases given 'm' remains unchanged and vice versa.

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A 50-gram ball is released from rest 80 m above the surface of the Earth. During the fall to the Earth, the total thermal energy of the ball and the air in the system increases by 15 J. Just before it hits the surface its speed is

Answers

Answer:

Speed of ball just before it hit the surface is 31.62 m/s .

Explanation:

Given :

Mass of ball , m = 50 g = 0.05 kg .

Height from which it falls , h = 80 m .

Thermal energy , E = 15 J .

Now , Initial energy of the system is :

$E_i=(mv^2)/(2)+mgh$

Here , initial velocity is zero .

Therefore , $E_i=mgh=40\ J$

Now , final energy of the system :

$E_f=(mv^2)/(2)+mg(0)+15\n\nE_f=(0.05* v^2)/(2)+15$

Since , no external force is applied .

Therefore , total energy of the system will be constant .

By conservation of energy :

$E_i=E_f\n40=(0.05v^2)/(2)+15\n\n25=(0.05v^2)/(2)\n\nv=31.62\ m/s$

Therefore , speed of ball just before it hit the surface is 31.62 m/s .

Final answer:

Using the principle of conservation of energy, the speed of the ball just before hitting the Earth's surface is found to be 79.2 m/s after accounting for the 15 J increase in thermal energy.

Explanation:

This question is concerned with the concept of conservation of energy, specifically the principles of potential and kinetic energy. When the ball is 80 meters above the Earth's surface, the total gravitational potential energy is m*g*h = 50g*9.8m/s²*80m = 39200 J (where m is mass, g is gravity, and h is height), and the kinetic energy is 0.

As the ball falls, its potential energy gets converted into kinetic energy, but we also know that the total thermal energy of the ball and the air in the system increases by 15 J. That means that not all the potential energy is converted into kinetic energy, 15 J is lost to thermal energy. So, the kinetic energy of the ball when it hits the Earth is 39200 J - 15 J = 39185 J.

Finally, we know that kinetic energy equals (1/2)*m*v², where v is the speed of the ball. Rearranging this formula to solve for v we get, v = sqrt((2*kinetic energy)/m) = sqrt((2*39185 J)/50g) = 79.2 m/s. So, just before the ball hits the surface, its speed is 79.2 m/s.

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determine exactly where to place a cart on the track so that it rolls down the track, flies through the air, and lands precisely at 1) the green line, 2) the red line, and 3) the blue line, on the first try.

Answers

Answer: i think you should place it on the red line

Explanation:

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