Which optical device can focus light to a point through reflection?

A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c

netineya [11]

Answer:

$E=\frac{\lambda}{2\pi r\epsilon_0}$

Explanation:

We are given that

Linear charge density of wire= $\lambda$

Radius of hollow cylinder=R

Net linear charge density of cylinder= $2\lambda$

We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R

By Gauss theorem

$\oint E.dS=\frac{q}{\epsilon_0}$

$q=\lambda L$

$E(2\pi rL)=\frac{L\lambda}{\epsilon_0}$

Where surface area of cylinder= $2\pi rL$

$E=\frac{\lambda}{2\pi r\epsilon_0}$

8 0

2 years ago

Sketch a position-time graph for a bear starting

Dmitrij [34]

Explanation:

hopefully that makes sense. the position doesn't change over the 5 seconds, meaning it's stopped but time still continues. then when the slope is negative this shows the bear's position becoming negative (backing up, changing direction).

3 0

2 years ago

What evidence contradicts the theory that moving two-ton stones was physically impossible for the egyptians, and must have been

SOVA2 [1]

Archaeological evidence shows that Egyptians worked together to build the pyramids. Remains of quarries and ancient tools suggest that large slabs were created from rock beds. The slabs were placed on sleds and pulled to the building site. To make this process easier, men most likely poured oil on the roadway. This process is depicted in tomb paintings that date back to 1900 BCE.

8 0

2 years ago

A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.

Aleonysh [2.5K]

Answer: $53.31\°$ East of North

Explanation:

We have the following data:

Speed of the wind from East to West: $6.68 m/s$

Speed of the bee relative to the air: $8.33 m/s$

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

$sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}$