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Gala2k [10]
3 years ago
11

The normal is a line perpendicular to the reflecting surface at the point of incidence.

Physics
1 answer:
ladessa [460]3 years ago
4 0

Answer:

True

Explanation:

The normal line is defined as the line which is perpendicular to the reflecting surface at the point where the incident ray meet with the reflecting surface.

The angle of incident is defined as the angle which is subtended by the incident ray with respect to the normal ray by consider the normal ray as the base line and angle is measured from the point where incident ray is incident on the reflecting surface of the mirror.

Similarly reflecting ray can be defined as the ray which is reflected after the incident of a ray and the angle subtended by the reflecting ray is measure with respect to normal ray by considering normal ray as a base line.

Therefore, the normal ray is the perpendicular line to the reflecting surface at the point of incidence.

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Correctly label the various structures of the membranous labyrinth of the ear.
Ronch [10]

Answer:

left side top to bottom:

- saccule

- utricle

- ampullae

- semicircular duct: anterior

- semicircular duct: lateral

- semicircular duct: posterior

right side top to bottom:

- cochlear duct

- spiral ganglion of cochlea

- cochlear nerve

- vestibular nerve

7 0
3 years ago
What method of separation would work best on a homogeneous mixture salt water?
Lera25 [3.4K]
The appropriate answer is c. evaporation. Evaporation is the process by which a liquid changes to a gas and in this case is the best way to remove the water from the homogenous mixture. The reaction can be sped up by heating the mixture. All the water will eventually evaporate leaving behind the sodium chloride crystals as a precipitate.
Filtration works best on mixtures that have precipitates and distillation is for separating liquids with different boiling points.
5 0
4 years ago
Read 2 more answers
A(n) _____ is a gap in the geologic record where some rock layers have been lost because of erosion.
Elenna [48]
I'm not sure, I think it's option A.

Let me know if I'm wrong!
4 0
4 years ago
For fully developed laminar pipe flow in a circular pipe, the velocity profile is u(r) = 2(1-r2 /R2 ) in m/s, where R is the inn
sdas [7]

Answer:

a) v_{max} = 2\ \textup{m/s}

b) v_{avg} = 1\ \textup{m/s}

c) Q = 1.256 × 10⁻³ m³/s

Explanation:

Given:

The velocity profile as:

u(r) = 2(1-\frac{r^2}{R^2} )

Now, the maximum velocity of the flow is obtained at the center of the pipe

i.e r = 0

thus,

v_{max}=u(0) = 2(1-\frac{0^2}{R^2} )

or

v_{max} = 2\ \textup{m/s}

Now,

v_{avg} = \frac{v_{max}}{2}\ \textup{m/s}

or

v_{avg} = \frac{2}}{2}\ \textup{m/s}

or

v_{avg} = 1\ \textup{m/s}

Now, the flow rate is given as:

Q = Area of cross-section of pipe × v_{avg}

or

Q = \frac{\pi D^2}{4}\times v_{avg}

or

Q = \frac{\pi 0.04^2}{4}\times 1

or

Q = 1.256 × 10⁻³ m³/s

7 0
3 years ago
Which data set has the largest standard deviation
AfilCa [17]

Answer:

<em>The data set marked as B has the largest standard deviation</em>

Explanation:

<u>Standard Deviation</u>

It's a number used to show how a set of measurements is spread out from the average value. A low standard deviation means that most of the values are close to the average. A high standard deviation means that the numbers are more spread out.

The formula for the standard deviation is

\displaystyle \sigma=\sqrt{\frac{\sum (x_i-\mu)^2}{n}}

Where x_i is the value of each measurement, n is the number of elements in the set, and \mu is the average or media of the values, defined as

\displaystyle \mu=\frac{\sum x_i}{n}

Let's analyze each set of data:

A.3,4,3,4,3,4,3

The average is

\displaystyle \mu=\frac{3+4+3+4+3+4+3}{7}=3.43

Computing the stardard deviation:

\sigma=\sqrt{\frac{(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2}{7}}

\sigma=0.5

B.1,6,3,15,4,12,8

The average is

\displaystyle \mu=\frac{1+6+3+15+4+12+8}{7}=7

Computing the stardard deviation:

\sigma=\sqrt{\frac{(1-7)^2+(6-7)^2+(3-7)^2+(15-7)^2+(4-7)^2+(12-7)^2+(8-7)^2}{7}}

\sigma=4.7

C. 20, 21,23,19,19,20,20

The average is

\displaystyle \mu=\frac{20+21+23+19+19+20+20}{7}=20.29

Computing the stardard deviation:

\sigma=\sqrt{\frac{(20-20.29)^2+(21-20.29)^2+(23-20.29)^2+(19-20.29)^2+(19-20.29)^2+(20-20.29)^2+(20-20.29)^2}{7}}

\sigma=1.3

D.12,14,13,14,12,13,12

The average is

\displaystyle \mu=\frac{12+14+13+14+12+13+12}{7}=12.86

Computing the stardard deviation:

\sigma=\sqrt{\frac{(12-12.86)^2+(14-12.86)^2+(13-12.86)^2+(14-12.86)^2+(12-12.86)^2+(13-12.86)^2+(12-12.86)^2}{7}}

\sigma=0.8

We can see the data set marked as B has the largest standard deviation

5 0
4 years ago
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