Ask question

Ask question

All categories

1 year ago

11

Which picture correctly shows the path of the reflected light rays given an object outside the focal point?Select one:a. Ab. Bc.

Cd. D

1 answer:

nignag [31]1 year ago

6 0

We will have that the graph that describes the scenario is given by graph B.

You might be interested in

A car is initially traveling at 12 m/s when the driver sees a yellow light ahead. He accelerates at a constant 7 m/s^2 for 6 s i

Vitek1552 [10]

Answer:

v = 54m/s

Explanation:

$a = \frac{v - u}{t}$

a = 7m/s²

u = 12m/s

t = 6s

7 = (v-12)/6

v - 12 = 42

v = 54m/s

(Correct me if i am wrong)

3 0

3 years ago

baherus [9]

Answer:

Mass is the correct answer.

Explanation:

Many drivers report a more positive handling response and a definite improvement when reducing unsprung mass. You want to keep unsprung weight to as little as possible. This minimizes the momentum and energies that your suspension has to counter. In effect, it can make your shocks more sensitive.

4 0

3 years ago

Read 2 more answers

An Olympic track runner starts from rest and has an acceleration of 2.4 m/s2 for 3.6 s, then has zero acceleration for the remai

rjkz [21]

Answer:

The runner's speed at the following times would remain 8.64 m/s.

Explanation:

Acceleration definition: Acceleration is rate of change in velocity of an object with respect to time.

In this case, after 3.6 seconds the acceleration is zero, it means that the velocity of the runner after 3.6 seconds is not changing and it will remain constant for the remainder of the race. Now, we have to find the velocity of the runner that he had after 3.6 seconds and that would be the runner's speed for the remainder of the race. For this we use first equation of motion.

First equation of motion: Vf = Vi + a×t

Vf stands for final velocity

Vi stands for initial velocity

a stands for acceleration

t stands for time

In the question, it is mentioned that the runner starts from rest so its initial velocity (Vi) will be 0 m/s.

The acceleration (a) is given as 2.4 m/s²

The time (t) is given as 3.6 s

Now put the values of Vi, a and t in first equation of motion

Vf = Vi + a×t

Vf = 0 + 2.4×3.6

Vf = 2.4×3.6

Vf = 8.64 m/s

So,the runner's speed at the following times would remain 8.64 m/s.

5 0

3 years ago

When the atmosphere is not quite clear, one may sometimes see colored circles concentric with the Sun or the Moon. These are gen

stealth61 [152]

Answer:

D) diffraction

Explanation:

Corona is an optical phenomenon produced by the diffraction of sunlight or moonlight, as light moves through water droplets in the atmosphere.

This phenomenon produces one or more diffuse concentric rings of light around the Sun or Moon, usually seen as colored circles.

Therefore, the explanation for these phenomena of colored concentric circles, sometimes seen with the Sun or the Moon involves diffraction.

5 0

3 years ago

Because of interstellar dust, astronomers can see at most about 5 kpc into the disk of the galaxy at visual wavelengths. What pe

NNADVOKAT [17]

Answer:

96%

Explanation

Let A the total area of the galaxy, is modeled as a disc:

A = πR^2 = π (25 kpc)^2

And let a be the area that astronomers are able to see:

a = πr^2 = π(5 kpc)^2

The percentage that can be seen is equal to 100 times the ratio of the areas, of the galaxy and the "visible" part:

P = 100 a/A = (5/25)^2 = 100/25 = 4%

Therefore, the percentage of the galaxy not included, i.e. not seen is:

(100-4)% = 96%

5 0

3 years ago