3 years ago

Seeing light coming out from a crack in the wall is what type of wave interaction?

Physics

1 answer:

Norma-Jean [14]3 years ago

3 0

Answer:

becuase its where a wave (light etc) goes through a narrow area or across an edge of an area

You might be interested in

A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 38 m/s when i

inysia [295]

Explanation:

look !

speed= 38m/s

start from rest= 0

5 0

3 years ago

There are two parts, answer both, please dont just say part 1 is the correct answer

nordsb [41]

Answer:

Im not sure

Explanation:

4 0

3 years ago

Read 2 more answers

How are electric motors and generators similar?

kotykmax [81]

Answer:

they both produce energy

Explanation:

7 0

3 years ago

What happens to the current supplied by the battery when you add an identical bulb in parallel to the original bulb?

Marysya12 [62]

Answer:

b. The current stays the same.

Explanation:

In the case given current is supplied by the battery to a bulb . Here, we should know that bulb also apply resistance to the flow of current .

Now, when an identical bulb is connected in parallel to the original bulb .

Therefore, both the resistance( bulb) are in parallel.

We know, when two resistance are in parallel , current through them is same and voltage is divided between them.

Therefore, in this case current stays same in the original bulb.

Hence, this is the required solution.

6 0

3 years ago

A hypothetical planet has a mass of one-half that of the earth and a radius of twice that of the earth.

Fed [463]

<h2>Option A is the correct answer.</h2>

Explanation:

Acceleration due to gravity

$g=\frac{GM}{r^2}$

G = 6.67 × 10⁻¹¹ m² kg⁻¹ s⁻²

Let mass of earth be M and radius of earth be r.

We have

$g=\frac{GM}{r^2}$

Now

A hypothetical planet has a mass of one-half that of the earth and a radius of twice that of the earth.

Mass of hypothetical planet, M' = M/2

Radius of hypothetical planet, r' = 2r

Substituting

$g'=\frac{GM'}{r'^2}\\\\g'=\frac{G\times \frac{M}{2}}{(2r)^2}\\\\g'=\frac{\frac{GM}{r^2}}{8}\\\\g'=\frac{g}{8}$

Option A is the correct answer.

6 0

3 years ago