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Nonamiya [84]
3 years ago
15

A gamma ray photon has an energy of 0.91 GeV. (1 GeV = 109 eV.) What is the wavelength of the gamma ray in fm? (1 fm = 10-15 m)?

Enter a number with 2 digits behind the decimal point.
Physics
1 answer:
anastassius [24]3 years ago
6 0

Answer:

\lambda=1.37 fm

Explanation:

The Planck Eistein relation, states that the energy of a photon is proportional to its frequency:

E=h\nu(1)

h is the Plank constant.The frequency of a photon is defined as the speed of light over its wavelength:

\nu=\frac{c}{\lambda}(2)

Replacing (2) in (1):

E=\frac{hc}{\lambda}\\\lambda=\frac{hc}{E}\\\lambda=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{0.91*10^{9}eV}\\\\\lambda=(1.37*10^{-15}m)*\frac{1fm}{10^{-15}m}\\\\\lambda=1.37 fm

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Masteriza [31]

Answer:

D, the lithosphere. (CRUST AND UPPER MANTLE)

Explanation:

A tectonic plate (also called lithospheric plate) is a massive, irregularly shaped slab of solid rock, generally composed of both continental and oceanic lithosphere. Plate size can vary greatly, from a few hundred to thousands of kilometers across; the Pacific and Antarctic Plates are among the largest. Plate thickness also varies greatly, ranging from less than 15 km for young oceanic lithosphere to about 200 km or more for ancient continental lithosphere (for example, the interior parts of North and South America).

Information found on:

<u>https://pubs.usgs.gov/gip/dynamic/tectonic.html#:~:text=A%20tectonic%20plate%20(also%20called,both%20continental%20and%20oceanic%20lithosphere.&text=Continental%20crust%20is%20composed%20of,such%20as%20quartz%20and%20feldspar.</u>

3 0
3 years ago
Read 2 more answers
Listed in the Item Bank are some important labels for sections of the image below. To find out more information about labels, so
Agata [3.3K]

Answer:

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6 0
4 years ago
An 800-N billboard worker stands on a 4.0-m scaffold supported by vertical ropes at each end. If the scaffold weighs 500 N and t
Arturiano [62]

Answer:

T = 850 N

Explanation:

given,

mass of billboard worker = 800 N

length of scaffold = 4 m

weight of the scaffold = 500 N

worker is standing at 1 m from one end.

Tension in the rope = ?

To calculate the tension in the string we have to balance the clockwise and counterclockwise moment of the system.

Weight of the worker and the weight of the scaffold will be in clockwise direction where as the tension will be in counterclockwise direction

now,

800 x 3 + 500 x 2 = T x 4

4T = 3400

T = 850 N

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5 0
4 years ago
Two campers in a national park hike from their cabin to the same spot on a lake, each taking a different path, as illustrated be
elixir [45]

Answer:

<em>Their displacement is 5 Km from the cabin.</em>

Explanation:

The displacement is define is the shortest distance between two points. It is the shortest distance of a line that joins two points together. It quantifies both the distance and direction of the net or total motion along a straight line from the initial position to the final position of the point trajectory. In this case, the straight line joining the cabin to the stream is line S which is equal to 5 Km

3 0
3 years ago
The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl
Ulleksa [173]

Hi there!

a)

We can begin by using the equation for energy density.

U = \frac{1}{2}\epsilon_0 E^2

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}

Creating a Gaussian surface being the lateral surface area of a cylinder:
A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}

Now, we can calculate the energy density using the equation:
U = \frac{1}{2} \epsilon_0 E^2

Plug in the expression for the electric field and solve.

U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}

b)

Now, we can integrate over the volume with respect to the radius.

Recall:
V = \pi r^2L \\\\dV = 2\pi rLdr

Now, we can take the integral of the above expression. Let:
r_i = inner cylinder radius

r_o = outer cylindrical shell inner radius

Total energy-field energy:

U = \int\limits^{r_o}_{r_i} {U_D} \, dV =   \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr

Plug in the equation for the electric field energy density and solve.

U =   \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\

Bring constants in front and integrate. Recall the following integration rule:
\int {\frac{1}{x}} \, dx  = ln(x) + C

Now, we can solve!

U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}

And here's our equation!

3 0
2 years ago
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