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2 years ago

Superman used X-ray vision to detect objects. Aside from a source of X-rays, what else would be needed for X-ray vision?

Physics

1 answer:

Arada [10]2 years ago

3 0

Answer:

an absorber of x-ray

Explanation:

To make x-ray detection/vision work, you will need at least two items: a source of x-ray and absorbed or x-ray.

The object you want to see itself doesn't have to be the source, but it has to absorb some of the rays instead. When doing a chest x-ray test, the medical employee will put your chest between absorber and source. The heart can absorb some of the rays so it will appear more white than lungs who made of air and won't absorb the rays.

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You are provided with three polarizers with filters making angles of (A) 90 ∘ , (B) 180 ∘ and (C) −45 ∘ with respect

irinina [24]

Answer:

Order of maximum transmission of the polarizer is A, C and B.

Solution:

As per the question:

For the first polarizer, the angle is quite insignificant:

(A) $90^{\circ}$ :

The light intensity after passing through the first polarizer is $I_{o}$ and this intensity does not depend on the angle of the polarizer.

Consider $90^{\circ}$ with the vertical, the intensity is given by:

$I = I_{o}cos^{2}90^{\circ}$

$I = I_{o}cos(2(45^{\circ})) = I_{o}(\frac{1+cos90^{\circ})}{2} = \frac{I_{o}}{2}$

(B) $180^{\circ}$ :

Suppose the second polarizer is $45^{\circ}$ with the vertical.

Now, intensity through the second polarizer:

$I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(- 45 - 90)$

$I' = \frac{I_{o}}{2}cos^{2}135^{\circ} = \frac{I_{o}}{4}$

Now, if we consider the second polarizer to be $180^{\circ}$ ,

$I' = \frac{I_{o}}{2}cos^{2}180^{\circ} = \frac{I_{o}}{2}cos^{2}(180^{\circ} - 90^{\circ}) = 0$

Now,

Intensity through the third polarizer, if it is $180^{\circ}$ with the vertical:

$I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(180 - (- 45))$

$I' = \frac{I_{o}}{8}$

5 0

2 years ago

In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water d

Alexxandr [17]

Answer:

a) a = g / 3

b) x (3.0) = 14.7 m

c) m (3.0) = 29.4 g

Explanation:

Given:-

- The following differential equation for (x) the distance a rain drop has fallen has the form:

$x*g = x * \frac{dv}{dt} + v^2$

- Where, v = Speed of the raindrop

- Proposed solution to given ODE:

v = a*t

Where, a = acceleration of raindrop

Find:-

(a) Using the proposed solution for v find the acceleration a.

(b) Find the distance the raindrop has fallen in t = 3.00 s.

Solution:-

- We know that acceleration (a) is the first derivative of velocity (v):

a = dv / dt ... Eq 1

- Similarly, we know that velocity (v) is the first derivative of displacement (x):

v = dx / dt , v = a*t ... proposed solution (Eq 2)

v .dt = dx = a*t . dt

- integrate both sides:

∫a*t . dt = ∫dt

x = 0.5*a*t^2 ... Eq 3

- Substitute Eq1 , 2 , 3 into the given ODE:

0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2

= 1.5 a^2 t^2

a = g / 3

- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:

x (t) = 0.5*a*t^2

x (t = 3.0) = 0.5*9.81*3^2 / 3

x (3.0) = 14.7 m

- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:

m (t) = k*x (t)

m (3.0) = 2.00*x(3.0)

m (3.0) = 2.00*14.7

m (3.0) = 29.4 g

6 0

2 years ago

The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu

Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

$E=\phi + K$

where

$E=\frac{hc}{\lambda}$ is the energy of the incident light, with h being the Planck constant, c being the speed of light, and $\lambda$ being the wavelength

$\phi$ is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

$\lambda=190 nm=1.9\cdot 10^{-7}m$ , so the energy of the incident light is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J$

Converting in electronvolts,

$E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV$

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

$\phi = E-K=6.5 eV-4.0 eV=2.5 eV$

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

$\lambda=510 nm=5.1\cdot 10^{-7} m$

So its energy is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J$

Converting in electronvolts,

$E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV$

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: $\phi = 4.5 eV$

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

$E=\phi+K=4.5 eV+2.7 eV=7.2 eV$

Then the copper is replaced with sodium, which has work function of

$\phi = 2.3 eV$

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

$K=E-\phi = 7.2 eV-2.3 eV=4.9 eV$

7 0

2 years ago

In aviation, it is helpful for pilots to know the cloud ceiling, which is the distance between the ground and lowest cloud. The

saw5 [17]

Answer:

$b = a \frac{cos(\alpha)}{sin(\alpha)} = a Cot(\alpha)$

Explanation:

In order to calculate the height reached by the beam when it hit the cloud, it is possible to use the following mathematical equations based on the Pythagoras' theorem:

$b = \sqrt[2]{H^{2} - a^{2}}$

$b = \sqrt[2]{H^{2} - H^{2} sin^{2} (\alpha)} \\b = H \sqrt[2]{1 - sin^{2} (\alpha)}$

And if:

$H = \frac{a}{sin(\alpha)}$

Then:

$b = \frac{a}{sin(\alpha)} \sqrt{1 - sin^{2}(\alpha)}\\b = \frac{a}{sin(\alpha)} \sqrt{1 - 1 + cos^{2}(\alpha)}\\$

$b = \frac{a}{sin(\alpha)} \sqrt{cos^{2}(\alpha)}\\b = a \frac{cos(\alpha)}{sin(\alpha)} = a Cot(\alpha)$

4 0

2 years ago

A 5.5Kg block is hanging from a rope that is wrapped around the outside of a 13Kg flywheel disk witha radius of 33cm that is hag

Sergeeva-Olga [200]

Answer:

$3.9m/s^{2}$

Explanation:

Using second law of motion

$a =\frac {m1 * g - \frac {T}{r}}{m1 + 0.5 * m2}$ where m1 is mass of block, m2 is mass of flywheel, g is acceleration due to gravity whose value is taken as $9.81 m/s^{2}$ , T is torque and r is radius

Substituting 5.5 Kg for m1, 13 Kg for m2, 0.33 m for r, 2.5 Nm for T we obtain

$a = \frac {5.5 \times 9.81 - \frac {2.5}{0.33}}{(5.5 + 0.5 \times13)}=3.9m/s^{2}$

8 0

2 years ago