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2 years ago

This ray diagram shows the image formed when a candle is placed in front of

Physics

1 answer:

weqwewe [10]2 years ago

8 0

Answer:

Explanation:

Virtual images are always right side up while real images are always upside down. Therefore, the is a virtual image that is smaller than the original.

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X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

5 0

3 years ago

A bullet is fired at an angle θ above the horizontal with an initial velocity of 800 m/s from the top of an 80 m high tower. Wha

Irina-Kira [14]

Answer:

pahingi po ng pic pls para masagutang kopo iyan

6 0

2 years ago

Study the following image of the moving car.

liq [111]

Potential energy is high and kinetic is equal i believe.

3 0

3 years ago

Read 2 more answers

You are riding in an elevator on the way to the

timofeeve [1]

Answer:

F = 104.832 N

Explanation:

given,

upward acceleration of the lift = 1.90 m/s²

mass of box containing new computer = 28 kg.

coefficient of friction = 0.32

magnitude of force = ?

box is moving at constant speed hence acceleration will be zero.

Now force acting due to lift moving upward =

F = μ m ( g + a )

F = 0.32 × 28 × ( 9.8 + 1.9 )

F = 104.832 N

hence, the force applied should be equal to 104.832 N

4 0

3 years ago

A car move at an initial velocity of 240m and reach at the final velocity of 540m in 8hours. calculate its acceleration.

disa [49]

Answer:

a = 0.01m/s²

Explanation:

V_f = V_0+a*t

V_f = Velocity final

V_0 = Velocity initial

a = acceleration

t = time

a = (V_f-V_0)/t

a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))

a = 0.01m/s²

6 0

2 years ago