The particle with charge q is now released and given a quick push; as a result, it acquires speed v. Eventually, this particle ends up at the center of the original square and is momentarily at rest. If the mass of this particle is m , what was its initial speed v ?.

Answers

Answer 1

Answer:

The given situation is illustrated below. A particle is released and given a quick push. As a result, it acquires a speed v. Eventually, this particle ends up at the center of the original square and is momentarily at rest. If the mass of this particle is m, the initial speed of the particle is $√((2qV/m).)$

To solve the problem, we need to apply the law of conservation of energy, which states that energy can neither be created nor destroyed; it can only be transformed from one form to another.
Initial potential energy = Final kinetic energy
The initial potential energy of the particle is given by
U = qV
where V is the potential difference between the corner and the center of the square.
At the center of the square, the potential energy is zero.
The final kinetic energy of the particle is given by
K = (1/2) mv^2
where m is the mass of the particle and v is its final velocity.
Since the particle is momentarily at rest at the center of the square, its final kinetic energy is zero.
Therefore, we have
qV = (1/2) mv^2
Solving for v, we get
v = $√((2qV/m).)$

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Answer:

D is not the a vector quantities

The speed of light is 3.00×108m/s. How long does it take for light to travel from Earth to the Moon and back again? Express your answer using two significant figures.

Answers

Answer:

v = 3×10^8 m/s

s= 384,400 km= 3.84×10^8 m/s

t = ?

v = s/t = 2s/t

t = 2s/v

t = (2×3.84×10^8) ÷ 3×10^8

t = 2.56 seconds

Explanation:

Earth's moon is the brightest object in our

night sky and the closest celestial body. Its

presence and proximity play a huge role in

making life possible here on Earth. The moon's gravitational pull stabilizes Earth's wobble on its axis, leading to a stable climate.

The moon's orbit around Earth is elliptical. At perigee — its closest approach — the moon comes as close as 225,623 miles (363,104 kilometers). At apogee — the farthest away it gets — the moon is 252,088 miles (405,696

km) from Earth. On average, the distance fromEarth to the moon is about 238,855 miles (384,400 km). According to NASA , "That means 30 Earth-sized planets could fit in between Earth and the moon."

Ultraviolet light is typically divided into three categories. UV-A, with wavelengths between 400 nm and 320 nm, has been linked with malignant melanomas. UV-B radiation, which is the primary cause of sunburn and other skin cancers, has wavelengths between 320 nm and 280 nm. Finally, the region known as UV-C extends to wavelengths of 100 nm. (a) Find the range of frequencies for UV-B radiation. (b) In which of these three categories does radiation with a frequency of 7.9 * 1014 Hz belong

Answers

Answer:

a) The UV-B has frequencies between $9.375x10^(14)Hz$ and $1.071x10^(15)Hz$

b) The radiation with a frequency of $7.9x10^(14)Hz$ belong to the UV-A category.

Explanation:

(a) Find the range of frequencies for UV-B radiation.

Ultraviolet light belongs to the electromagnetic spectrum, which distributes radiation along it in order of different frequencies or wavelengths.

Higher frequencies:

Gamma ray

X ray

Ultraviolet rays

Visible region

Lower frequencies:

Infrared

Microwave

Radio waves

That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it. Any of those radiations will have a speed of $3x10^{8]m/s$ in vacuum.

The velocity of a wave can be determined by means of the following equation:

$c = \nu \cdot \lambda$ (1)

Where c is the speed of light, $\nu$ is the frequency and $\lambda$ is the wavelength.

Then, from equation 1 the frequency can be isolated.

$\nu = (c)/(\lambda)$ (2)

Before using equation 2 to determine the range of UV-B it is necessary to express $\lambda$ in units of meters in order to match with the units from c.

$\lambda = 320nm . (1m)/(1x10^(9)nm)$ ⇒ $3.2x10^(-7)m$

$\lambda = 280nm . (1m)/(1x10^(9)nm)$ ⇒ $2.8x10^(-7)m$

$\nu = (3x10^(8)m/s)/(3.2x10^(-7)m)$

$\nu = 9.375x10^(14)s^(-1)$

$\nu = 9.375x10^(14)Hz$

$\nu = (3x10^(8)m/s)/(2.8x10^(-7)m)$

$\nu = 1.071x10^(15)Hz$

Hence, the UV-B has frequencies between $9.375x10^(14)Hz$ and $1.071x10^(15)Hz$

(b) In which of these three categories does radiation with a frequency of $7.9x10^(14)Hz$ belong.

The same approach followed in part A will be used to answer part B.

Case for UV-A:

$\lambda = 400nm . (1m)/(1x10^(9)nm)$ ⇒ $4x10^(-7)m$

$\nu = (3x10^(8)m/s)/(4x10^(-7)m)$

$\nu = 7.5x10^(14)s^(-1)$

$\nu = 7.5x10^(14)Hz$

Hence, the UV-A has frequencies between $7.5x10^(14)Hz$ and $9.375x10^(14)Hz$ .

Therefore, the radiation with a frequency of $7.9x10^(14)Hz$ belongs to UV-A category.

Complete the calculations for total magnification produced by various combinations of the eyepiece and objective lenses. You may assume that the magnification for the eyepiece is 10X for each question. 1. When the scanning (4X) objective is used the total magnification will be:________
2. When the low power (10X) objective is used the total magnification will be:________
3. When the high power (40X) objective is used the total magnification will be:________
4. When the oil immersion (100X) objective is used the total magnification will be:_________

Answers

Answer:

a) m_ttoal = 40x, b) m_total = 100X, c) m_total = 400X,

d) m_total = 1000 X

Explanation:

La magnificación o aumentos es el incremento de del tamaño de la imagen con respecto al tamaño original del objeto, en la mayoria del os sistema optico la magnificacion total es el producoto de la magnificación del objetivo por la magnificación del ocular

m_total = m_ objetivo * m=ocular

apliquemos esto a nuestro caso

1) m_total = 4 x * 10 x

m_ttoal = 40x

2) m_total = 10X * 10X

m_total = 100X

3)mtotal = 40X * 10X

m_total = 400X

4) m _totla = 100x * 10 X

m_total = 1000 X

en este ultimo caso para magnificación grandes es decalcificar el objeto

The total magnification produced by different combinations of eyepiece and objective lenses in a microscope.

1. When the scanning (4X) objective is used, the total magnification will be 40X because the eyepiece magnification is 10X and the objective magnification is 4X.

2. When the low power (10X) objective is used, the total magnification will be 100X because the eyepiece magnification is 10X and the objective magnification is 10X.

3. When the high power (40X) objective is used, the total magnification will be 400X because the eyepiece magnification is 10X and the objective magnification is 40X.

4. When the oil immersion (100X) objective is used, the total magnification will be 1000X because the eyepiece magnification is 10X and the objective magnification is 100X.

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If R = 12 cm, M = 430 g, and m = 60 g , find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (Treat the pulley as a uniform disk.)

Answers

Answer:

Explanation:

Given

Radius of Pulley r=12 cm

mass of block m=60 gm

mass of Pulley M=430 gm

Block descend h=50 cm

Applying Conservation of Energy

Potential Energy of block convert to rotational Energy of pulley and kinetic energy of block

i.e.

$mgh=(1)/(2)I\omega ^2+(1)/(2)mv^2$

where I=moment of inertia

$I=mr^2$

and for rolling $\omega =(v)/(r)$

$mgh=(1)/(2)Mv^2+(1)/(2)mv^2$

$v^2=(2mgh)/(m+M)$

$v=\sqrt{(2mgh)/(m+M)}$

$v=\sqrt{(2* 60* 9.8* 0.5)/(430+60)}$

$v=\sqrt{(60* 9.8)/(490)}$

$v=√(1.2)$

$v=1.095 m/s$

A circular loop of wire has radius 7.80cm . A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 2.03�10?2W/m2 , and the wavelength of the wave is 6.20m .What is the maximum emf induced in the loop? Express your answer with the appropriate units.I stumbled through the formulas I do know for EMF but cant seem to figure out how to get the right answer. Please help and provide explanation! Thanks

Answers

Answer:

fem = - 4.50 10²² V

Explanation:

For the solution of this problem we must use the equation of the induced electromotive force or Faraday's law

E = - d Φ $._(B)$ / dt = d (BA cos θ) dt

In this case they tell us that the magnetic field is perpendicular to the plane of the loop, as the normal to the surface of the loop is in the direction of the radius, the angle between the field and this normal is zero, so cos 0º = 1. The area of the loop is constant, with this the equation is

E = - A dB / dt (1)

To find field B, we have the relationships of electromagnetic waves

E = c B

The intensity or poynting vector for the wave is described by the equation

S = I = 1 / μ₀ E x B = 1 /μ₀ E B

We replace

I = 1 /μ₀ (cB) B = c /μ₀ B²

This is the instantaneous intensity.

B = √ (μ₀ I /c)

We substitute in equation 1

E = - A μ₀/c d I / dt

With the maximum value we are asked to change it derived from variations

E = -A c/μ₀ ΔI / Δt

It remains to find the time of the variation. Let's use the equation

c = λ f = λ / T

T = λ / c

T = 6.20 / 3 10⁸

T = 2.06 10⁻⁸ s

We already have all the values to calculate the fem

fem = - π r² c/μ₀ ΔI/Δt

fem = - (π 0.078²) (3 10⁸/(4π 10⁻⁷) (2.03 10² -0) / (2.06 10⁻⁸ - 0)

fem = - 4.50 10²² V

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