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3 years ago

Explain resolution of Force

Physics

1 answer:

Margarita [4]3 years ago

7 0

Answer:

it is defined as splitting up the given force into a number of components, without changing its effects on the body is called resolution of forces. A force is generally resolved along with two mutually perpendicular directions.

Explanation:

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X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{

dedylja [7]

<h2>Answer: 37.937 keV</h2>

Explanation:

<u>Photons have momentum</u>, this was proved by he American physicist Arthur H. Compton after his experiments related to the <u>scattering of photons from electrons</u> (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

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

Where:

$\lambda_{c}=2.43(10)^{-12} m$ 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.

$\theta=30\°$ the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at $30\°$ :

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))$ (2)

$\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m$ (3)

Now, the initial energy $E_{o}=400keV=400(10)^{3}eV$ of the photon is given by:

$E_{o}=\frac{h.c}{\lambda_{o}}$ (4)

From this equation (4) we can find the value of $\lambda_{o}$ :

$\lambda_{o}=\frac{h.c}{E_{o}}$ (5)

$\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}$

$\lambda_{o}=3.102(10)^{-12}m$ (6)

Knowing the value of $\Delta \lambda$ and $\lambda_{o}$ , let's find $\lambda'$ :

$\Delta \lambda=\lambda' - \lambda_{o}$

Then:

$\lambda'=\Delta \lambda+\lambda_{o}$ (7)

$\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m$

$\lambda'=3.427(10)^{-12}m$ (8)

Knowing the wavelength of the scattered photon $\lambda'$ , we can find its energy $E'$ :

$E'=\frac{h.c}{\lambda'}$ (9)

$E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}$

$E'=362.063keV$ (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron $E_{e}$ , we have to calculate all the energy lost by the photon, which is:

$E_{e}=E_{o}-E'$ (11)

$E_{e}=400keV-362.063keV$

Finally we obtain the energy of the recoiling electron:

$E_{e}=37.937keV$

5 0

4 years ago

A projectile is launched on the Earth with a certain initial velocity and moves without air resistance. Another projectile is la

WINSTONCH [101]

With smaller gravitational forces and therefor less vertical acceleration, the projectile launched on the moon ... with the same initial speed and direction ...

-- climbs faster,

-- spends more time climbing,

-- reaches a higher peak,

-- falls slower,

-- spends more time falling, and

-- covers more horizontal distance

than the projectile launched on the Earth.

This is not because of air resistance. It would be true even if there were no air resistance on the Earth. It's entirely a gravity thing.

4 0

4 years ago

What are the units of density

Bad White [126]

Answer:

SI unit of density - Kilogram per cubic metre (Kg/m³)

CGS unit of density - gram per cubic centimetre (g/cm³)

5 0

3 years ago

The average bulk resistivity of the human body (apart from surface resistance of the skin) is about 5.0 Ω⋅m. The conducting path

Varvara68 [4.7K]

Answer:

A) R = 1019.11 Ω

B) V = 101.91 V

Explanation:

a) The resistance is given by the formula;

R = ρL/A

Where;

ρ is bulk density = 5.0 Ωm

L is length of cylinder = 1.6m

A is area = (π/4)D² = (π/4) x 0.1² = 0.007854 m²

Plugging in the relevant values, we obtain ;

R = (5 Ωm x 1.6m)/(0.007854 m²)

R = 1019.11 Ω

b) potential difference is given by;

V = RI

Thus, plugging in relevant values, we obtain;

V = 1019.11 Ω ∙ 0.1A

V = 101.91 V

7 0

3 years ago

g A 10kg weight is suspended from the ceiling by a spring. The weight-spring system is at equilibrium with the bottom of the wei

VladimirAG [237]

Answer:

Reverses its direction of travel precisely as it reaches the eggs

Explanation:

At the top of the motion: The spring is stretched the least, so, the potential energy of the spring is at a minimum. The mass is lifted as high as it will go, so, the potential energy due to gravity is at a maximum.

When the spring is now at the maximum, now, the maximum potential at the maximum height is

Equal to the energy stored in spring

And the energy stored in spring.

The net force on the object can be described by Hooke’s law, and so the object undergoes simple harmonic motion. Note that the initial position has the vertical displacement at its maximum value X; v is initially zero and then negative as the object moves down; and the initial acceleration is negative, back toward the equilibrium position and becomes zero at that point.

So this apply to the body given, when it get to the egg and it is released the weight will move upward 1m above the equilibrium point and it will return downward 1m below the equilibrium point, that is reverses its direction of travel precisely as it reaches the eggs.

6 0

3 years ago

Explain resolution of Force​

Explain resolution of Force