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vesna_86 [32]
3 years ago
8

When light strikes an opaque material, which of the following accurately describes what happens to the light rays? Some of the l

ight passes through it but that light is quickly scattered; some of the light is absorbed as heat and some is reflected off the surface Most of the light passes through it, but some is also reflected and refracted None of the light passes through it; some of the light is absorbed as heat but most is reflected off the surface Half of the light rays pass through heating the substance and the other half are reflected off the surface casting a shadow
Physics
2 answers:
ExtremeBDS [4]3 years ago
8 0
<span>None of the light passes through it; some of the light is absorbed as heat but most is reflected off the surface.  This is how you see </span>objects.  reflected light from them hits your eye.  (Opaque means not transparent)
statuscvo [17]3 years ago
8 0

Answer:

None of the light passes through it some of the light is absorbed as heat but most is reflected off the surface

Explanation:

The materials are of three types i.e. transparent, translucent and opaque materials. Opaque materials are the materials that do not allow the light rays to pass through it. Some of the examples of opaque materials are wood, metals, dark colored plastics, bricks etc.

Usually light some many phenomenons like scattering, absorption, reflection etc.

Hence, the correct statement about opaque material is " None of the light passes through it, some of the light is absorbed as heat but most is reflected off the surface".

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Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce
lutik1710 [3]

Answer:

speed of electrons = 3.25 × 10^{7} m/s

acceleration in term g is 3.9 × 10^{17} g.

radius of circular orbit is 2.76 × 10^{-4} m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = \sqrt{\frac{2qV}{m}}

v = \sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}

v = 3.25 × 10^{7} m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = \frac{Bqv}{m}  

a =  \frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}

a = 3.82 × 10^{18} m/s²

and acceleration in term g

a = \frac{3.82\times 10^{18}}{9.81}  

a = 3.9 × 10^{17} g

acceleration in term g is 3.9 × 10^{17} g.

and

electron moving in circular orbit has centripetal force

F = \frac{mv^2}{r}  

Bqv = \frac{mv^2}{r}  

r = \frac{mv}{Bq}  

r = \frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}  

r = 2.76 × 10^{-4} m

radius of circular orbit is 2.76 × 10^{-4} m

8 0
3 years ago
Which of the following elements is in the same period as phosphorus?
Arte-miy333 [17]
The answer is B. magnesium I am pretty sure
5 0
3 years ago
We dont see objects. We see the light ____ off objects.
Romashka [77]
The answer is BBBBBBBBB
4 0
3 years ago
A commuter train passes a passenger platform at a constant speed of 40.4 m/s. The train horn is sounded at its characteristic fr
mihalych1998 [28]

(a) -83.6 Hz

Due to the Doppler effect, the frequency of the sound of the train horn appears shifted to the observer at rest, according to the formula:

f' = (\frac{v}{v\pm v_s})f

where

f' is the apparent frequency

v = 343 m/s is the speed of sound

v_s is the velocity of the source of the sound (positive if the source is moving away from the observer, negative if it is moving towards the observer)

f is the original frequency of the sound

Here we have

f = 350 Hz

When the train is approaching, we have

v_s = -40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s - 40.4 m/s})(350 Hz)=396.7 Hz

When the train has passed the platform, we have

v_s = +40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s + 40.4 m/s})(350 Hz)=313.1 Hz

Therefore the overall shift in frequency is

\Delta f = 313.1 Hz - 396.7 Hz = -83.6 Hz

And the negative sign means the frequency has decreased.

(b) 0.865 m

The wavelength and the frequency of a wave are related by the equation

v=\lambda f

where

v is the speed of the wave

\lambda is the wavelength

f is the frequency

When the train is approaching the platform, we have

v = 343 m/s (speed of sound)

f = f' = 396.7 Hz (apparent frequency)

Therefore the wavelength detected by a person on the platform is

\lambda' = \frac{v}{f'}=\frac{343 m/s}{396.7 Hz}=0.865m

5 0
3 years ago
Does anybody know how to answer any of these questions? Any of them without a check mark.
aivan3 [116]

Explanation:

sorry I can't help u right now

3 0
3 years ago
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