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3 years ago

The line that is drawn perpendicular to the point at which a wave intersects a boundary is know as the

1 answer:

fomenos3 years ago

5 0

<span>The line that is drawn perpendicular to the point at which a wave intersects a boundary is know as the Normal .

When the normal is drawn, the incident ray makes an angle with it known as the angle of incidence and the reflected ray makes an angle with it known as the angle of incidence. These angles are always equal.

The refracted ray makes an angle with the normal known as angle of refraction. The sin of angle of incidence to the sin of angle of refraction is called the refractive index( </span>μ= <span>sin i / sin r) .

hope all of it helps you!</span>

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A block of aluminum at a temperature of T1 = 32 degrees C has a mass of m1 = 12 kg. It is brought into contact with another bloc

Mariulka [41]

Answer: $T=21.33 ^{\circ}$

Explanation:

Given

mass of First Block $m_1=12 kg$

Temperature $T_1=32^{\circ}$

mass of second block $m_2=0.5 m_1=6 kg$

Temperature $T_2=9^{\circ}$

Heat capacity of aluminium c=899 J/kg-K

Final Temperature acquired by both blocks at steady state

Heat loss first block =Heat gain by second block

$m_1\times c\times (32-T)=m_2\times c\times (T-9)$

$12\times 899\times (32-T)=6\times 899\times (T-9)$

$2\times 32=3T$

$T=\frac{64}{3}=21.33 ^{\circ}$

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3 years ago

A bystander observes the musicians heading toward each other. When musician #1 is 100 m away, the intensity is 1.24 x 10-8 W/m^2

777dan777 [17]

Explanation:

Given that,

Distance 1, r = 100 m

Intensity, $I_1=1.24\times 10^{-8}\ W/m^2$

If distance 2, r' = 25 m

We need to find the intensity and the intensity level at 25 meters. Intensity and a distance r is given by :

$I=\dfrac{P}{4\pi r^2}$ .........(1)

Let I' is the intensity at r'. So,

$I'=\dfrac{P}{4\pi r'^2}$ ............(2)

From equation (1) and (2) :

$I'=\dfrac{Ir}{r'^2}$

$I'=\dfrac{1.24\times 10^{-8}\times 100}{25^2}$

$I'=1.98\times 10^{-9}\ W/m^2$

Intensity level is given by :

$dB=10\ log(\dfrac{I'}{I_o})$ , $I_o=10^{-12}\ W/m^2$

$dB=10\ log(\dfrac{1.98\times 10^{-9}}{10^{-12}})$

dB = 32.96 dB

Hence, this is the required solution.

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3 years ago

The wavelength of a wave on a string is 1.2 meters. If the speed of the wave is 60 meters/second, what is its frequency?

maria [59]

F=v/wavelength f=1.2/60=50Hz.

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3 years ago

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The hubbles telescopes orbit is 5.6 x10 ^5 meters above earths suface. the telescope has a mass os 1.1 x10^4 kilograms. earth ex

Andru [333]

(3) 8.3 N/kg. The gravitational field strength at a point is the force per unit mass exerted on a mass placed at that point. So at the point where the Hubble telescope is, it is (9.1 x 10^4)N/(1.1 x 10^4 kg) = 8.3 N/kg

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3 years ago

What are the two types of electrical circuits

uysha [10]

Series Circuit
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Parallel Circuit
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3 0

3 years ago

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