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3 years ago

How can we represent a narrow beam of light?

2 answers:

tatiyna3 years ago

8 0

Answer:It is easy to imagine representing a narrow beam of light by a collection of parallel arrows—a bundle of rays. As the beam of light moves from one medium to another, reflects off surfaces, disperses, or comes to a focus, the bundle of rays traces the beam's progress in a simple geometrical manner.

Explanation:

Yanka [14]3 years ago

7 0

Answer:

It is easy to imagine representing a narrow beam of light by a collection of parallel arrows—a bundle of rays. As the beam of light moves from one medium to another, reflects off surfaces, disperses, or comes to a focus, the bundle of rays traces the beam's progress in a simple geometrical manner.

I hope it's helpful!

You might be interested in

Between a piece of paper and a rock, which will fall the fastest and why?

AysviL [449]

Answer:

The Rock

Explanation:

The paper is very thin and light so it has more air reisistance which will cause it to fall slower than the rock that has a larger density.

7 0

3 years ago

0.001225 kg/L x 720 000 000L =?

I am Lyosha [343]

Answer:

0.001225 kg/L × 720 000 000 L = 882000 kg

Explanation:

Given:

The equation to solve is given as:

0.001225 kg/L × 720 000 000 L = ?

Let us write each term of the product in terms of power of 10.

As 0.001225 has 6 digits after the decimal place, therefore, we use the exponent 6 for 10 and the sign is negative. This gives,

$0.001225\ kg/L = 1225\times 10^{-6}\ kg/L$

Now, for 720000000 L there are 6 zeros after 720. So, we use exponent 6 but with a positive sign. This gives,

$720000000\ L=720\times 10^{6}\ L$

Now, finding the product, we get:

$0.001225\ kg/L\times 720000000\ L\\\\=1225\times 10^{-6}\ kg/L\times 720\times 10^6\ L\\\\=(1225\times 720)\times (10^6\times 10^{-6})\ (\frac{kg}{L}\times L)\\\\=882000\times 10^{6-6}\ kg\\\\=882000\times 10^0\ kg\\\\=882000\times 1\ kg\\\\=882000\ kg$

Therefore, the product is equal to:

0.001225 kg/L × 720 000 000 L = 882000 kg

6 0

3 years ago

After watching this video, Blake, a student in an introductory physics class, makes the following claim: The acceleration and ve

Maru [420]

Answer:

Please see below as the answer is self- explanatory.

Explanation:

Any time that an object changes direction (from leftward to rightwward, or from upward to downward) the velocity must be zero just for one instant, when is on the verge of changing the direction.
This is needed because velocity changes as a continuous function of time, so it needs to cross the t-axis when passing from positive to negative or vice versa.
However, the claim that in the moment that velocity is zero, the acceleration is also zero, is false.
Due to acceleration is the rate of change of velocity, and velocity is a vector, this means at any time there is a change of direction, there is an acceleration that is non-zero.
For example, when an object that has been thrown upward, reaches to its maximum height, just one instant before starting to fall, the velocity becomes zero, but the acceleration (which causes the object to fall) is non-zero, due to it's the acceleration due to gravity.

8 0

3 years ago

According to coulombs law, what will happen to the force between two charged particles if the magnitude are increased by 6 times

seropon [69]

Answer: The electrostatic force will be the same

Explanation:

According to Coulomb's Law, when two electrically charged bodies come closer, appears a force that attracts or repels them, depending on the sign of the charges of this two bodies or particles.

In this sense, this law states the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Being $K$ is a proportionality constant.

Now, if each $q_{1}$ and $q_{2}$ are increased by 6, and the distance between them as well, we will have the following:

$F_{E}= K\frac{6 q_{1}. 6 q_{2}}{(6d)^{2}}$ (2)

$F_{E}= 36 K\frac{q_{1}. q_{2}}{36d^{2}}$ (3)

Simplifying:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (4)

Comparing (1) with (4) we can see the electrostatic force is the same.

4 0

4 years ago

(1 point) At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 22 knots and ship B is sailing nort

Veronika [31]

Answer:28.8 knots

Explanation:

The ships are moving as the sides of a right triangle. Thus, Pyhogorean theorem will be useful in the following steps. Next, we have to know that the rate of change in distance, which is called velocity, can be described in terms of derivatives.

First, we have to calculate the distances covered by the ships from noon to 6 PM. In 6 hours, ship A moved 22*6=132 nautical mile. However, their first distance was 10 nautical miles, so 132+10=142 miles is the equivalent of A's displacement. For B, the distance travelled is 19*6=114 miles. From now on, A=142 miles and B=114 miles.

The distance between them is described with Pythogorean theorem, which is $D=\sqrt{A^{2} +B^{2} }$ and when we replace the values A and D, we find Distance (D) to be 182 miles.

Now, let's make the notations clear. The velocity of A and B is notated as $\frac{dA}{dt}$ and $\frac{dB}{dt}$ . The rate of change of distance is also notated as $\frac{dD}{dt}$ . Now, we have to find $\frac{dD}{dt}$ from the Pythogorean theorem. If we derive the Pythogorean expression $D=\sqrt{A^{2} +B^{2} }$ , we would have:

$\frac{dD}{dt} =\frac{1}{2} *(A^{2} +B^{2} )^{-1/2} *(2*A*\frac{dA}{dt} + 2*B*\frac{dB}{dt} )$

The derivation here includes chain rule and derives the interior parts of the parenthesis. When we insert distances for A and B and velocities for derivation notations, the formula becomes:

$\frac{dC}{dt} =\frac{1}{2}*(142^{2} +114^{2})^{-\frac{1}{2} }*(2*142*22 + 2*114*19)$ and the answer is 28.6 knots.

6 0

3 years ago