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4 years ago

A given 60uc piont charge located at origion

2 answers:

7 0

For my first part it like that as we all know that flux density is the charge per unt area here charge is 60uc so divide 60uc by (4*pi*r2) we get D= 7.06*10^-5 c/m2. NOW given is portion with r=26cm theta= 0 to pi/2 and phi = 0 to pi/2 calculate required region area with formula =double integral(r^2sintheta dtheta dphi) we get =.106m^2 now multiply D*Required region we get 7.5uc M I RIGHT PLZ tell me

Morgarella [4.7K]4 years ago

6 0

Answer:

flux density is the charge per unt area here charge is 60uc so divide 60uc by (4*pi*r2) we get D= 7.06*10^-5 c/m2. NOW given is portion with r=26cm theta= 0 to pi/2 and phi = 0 to pi/2 calculate required region area with formula =double integral(r^2sintheta dtheta dphi) we get =.106m^2 now multiply D*Required region we get 7.5uc

Explanation:

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Consider Compton Scattering with visible light.A photon with wavelength 500nm scatters backward(theta=180degree) from a free ele

JulijaS [17]

Answer: $4.86(10)^{-12}m$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda'=500 nm=500(10)^{-9} m$ is the wavelength of the scattered photon

$\lambda_{o}$ is the wavelength of the incident photon

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=180\°$ the angle between incident phhoton and the scatered photon.

$\Delta \lambda=2.43(10)^{-12} m (1-cos(180\°))$ (2)

$\Delta \lambda=4.86(10)^{-12}m$ (3) This is the shift in wavelength

5 0

3 years ago

Particle 1 has mass 4.6 kg and is on the x-axis at x = 5.7 m. Particle 2 has mass 7.2 kg and is on the y-axis at y = 4.2 m. Part

Iteru [2.4K]

To solve this problem it is necessary to apply the concepts related to the Gravitational Force, for this purpose it is understood that the gravitational force is described as

$F_g = \frac{Gm_1m_2}{r^2}$

Where,

G = Gravitational Universal Force

$m_i =$ Mass of each object

To solve this problem it is necessary to divide the gravitational force (x, y) into the required components and then use the tangent to find the angle generated between both components.

Our values are given as,

$m_1 =4.6 kg\\m_2 = 7.2 kg\\m_3 = 2.6 kg\\r_1 = 5.7 m\\r_2 = 4.2 m$

Applying the previous equation at X-Axis,

$F_x = \frac{Gm_1m_3}{R_{1}^2}\\F_x = \frac{6.67*10^{-11}*4.6*2.6}{5.7^2}\\F_x = 2.46*10^{-11}N$

Applying the previous equation at Y-Axis,

$F_y = \frac{Gm_2m_3}{R_2^2}\\F_y = \frac{6.67*10^{-11}*7.2*2.6}{4.2^2}\\F_y = 7.08*10^{-11} N$

Therefore the angle can be calculated as,

$tan\theta = \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{7.08*10^{-11}}{2.46*10^{-11}}\\\theta = 71\°$

Then in the measure contrary to the hands of the clock the Force in the particle 3 is in between the positive direction of the X and the negative direction of the Y at 71 ° from the positive x-axis.

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4 years ago

Khalad is measuring the amplitude of a wave. What can be known about this wave?

GalinKa [24]

It is B) transverse wave

6 0

3 years ago

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Why is the temperature of the liquid in the flask on the previous page measured when the liquid in the thermometer has stopped r

Flura [38]

when heated, the molecules of the liquid in the thermometer move faster, causing them to get a little further apart. this results in movement up the thermometer. when cooled, the molecules of the liquid in the thermometer move slower, causing them to get a little closer together.

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3 years ago

Define polarization in reference in the simple cell

vampirchik [111]

Answer:polarization pertains to the act or process of producing a positive electrical charge and a negative electrical charge such that between a nerve cell internal electrical charge..................

Explanation:

7 0

4 years ago