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I am Lyosha [343]
3 years ago
7

Which statement is true for light passing into a medium that is less optically dense than the first medium through which it pass

ed?
The index of refraction of the second medium is higher.


The index of reflection of the second medium is lower.


The index of reflection of the second medium is higher.


The index of refraction of the second medium is lower.
Physics
2 answers:
BigorU [14]3 years ago
5 0

Answer:

The correct option is;

The index of refraction of the second medium is lower

Explanation:

The index of refraction of a material indicates the magnitude of the optical density of a material. The index of refraction or the refractive index, n, are indices (ratio) of the speed of light through an optically dense medium relative to the speed of light through a vacuum.

The definition of the refractive index is the number of times light travelling through a medium would be slower than light travelling through vacuum

Therefore, the index of refraction of a second medium that is less optically dense than a first medium from which light originates and travels through it would be lower than the index of refraction of the first medium

AleksAgata [21]3 years ago
5 0

Answer:

D

Explanation:

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What does wave speed have to do with all SONAR AND RADAR technologies?
Elena L [17]

Explanation:

1) Radar uses radio waves, which are a type of electromagnetic energy. Sonar uses the echo principle by sending out sound waves underwater or through the human body to locate objects. Sound waves are a type of acoustic energy. Because of the different type of energy used in radar and sonar, each has its own applications.

2)Radar systems operate using radio waves primarily in air, while sonar systems operate using sound waves primarily in water (Minkoff, 1991). Despite the difference in medium, similarities in the principles of radar and sonar can frequently result in technological convergence.

4 0
3 years ago
Read 2 more answers
2. What is the water pressure at a depth of 24 m in a lake? [Density of water, p = 1 000 kg m- and gravitational acceleration, g
Andrei [34K]

Answer:

Pressure = ρgh

pressure (p) is measured in pascals (Pa)

density (ρ) is measured in kilograms per metre cubed (kg/m3)

The fore of gravitational field strength (g) is measured in N/kg or m/s 2

height of column (h) is measured in metres (m)

Answer = 235,200 Pa

Explanation:

Pressure = ρgh

Pressure = 1,000 x 9.8 x 24

Pressure = 235,200 Pa

4 0
2 years ago
4) A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.
VMariaS [17]

Answer:

(a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

Explanation:

Given that,

A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.

Suppose, the distance between 40 yard line and 25 yard line is 20 yard.

(a). We need to calculate their speed during that run

Using formula of speed

v=\dfrac{d}{t}

Where. d = distance

t = time

Put the value into the formula

v=\dfrac{18.288}{2}

v=10\ m/sduring that run

(b). We need to calculate their velocity

Using formula of speed

v=\dfrac{\Delta d}{\Delta t}

Put the value into the formula

v=\dfrac{22.86-36.58}{2}

v=-6.86\ m/s

Negative sign shows the direction of motion.

(c). If they kept running at that velocity for another 1.3 seconds,

We need to calculate the final position

Using formula of position

d=vt

Put the value into the formula

d=6.86\times1.3

d=8.91\ m

Hence, (a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

8 0
3 years ago
Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same tim
blagie [28]

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

x=x_{0}+V t \\ x=\frac{1}{2}+V t

The distance between the police car and the intersection is,

y=y_{0}+V t

y=\frac{1}{2}-40 t

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })

z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)

The rate of change is,

2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)

2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots

Now finding z when t=0, from (1) we have

z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}

z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071

The officer's radar gun indicates 25 mph pointed at the other car then, \frac{d z}{d t}=25 when t=0, from

From (2) we get

2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)

2(0.7071)(25)=V+2 V^{2}(0)-40

35.36=V-40

V=35.36+40=75.36

Hence the speed of the car is 75.36 mph

7 0
3 years ago
Explain<br> (0) what is meant by regeneration,<br> (ii) why an analogue signal cannot be regenerated
Kamila [148]

Explanation:

<h3>1.) Regeneration is the natural process of replacing or restoring damaged or missing cells, tissues, organs, and even entire body parts to full function in plants and animals.</h3>

2.) When noise is added to analogue signals, it usually sounds like background hiss. Such noise can not be removed so the original clean signal can not be re-created or re-generated.

6 0
3 years ago
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