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Elena-2011 [213]
3 years ago
6

Which describe reflection? Check all that apply.

Physics
2 answers:
BigorU [14]3 years ago
8 0

Answer:

<h2>Light bounces off a boundary.</h2><h2>Light direction is determined using the law of reflection.</h2>

Explanation:

In physics, reflection is studied using light rays. Specifically, it's the process of bouncing off an object surface, that is, the light hits a surface and "move back".

This subject is studied in geomtric optics, because we use the geometry tools to understand, analyse and predict the behaviour of a light beam when hits an object.

Reflection is studied applying the Law of Reflection which statest that the angle of incidence is congruent with the angle of reflection regarding a normal axis. The image attached shows this law.

Therefore, the correct answer are the first and last choice, because as we said before, reflection is the light action of bouncing off a boundary or a object surface, if light doesn't bounce off and instead goes through, that's called refraction. Also, we saw that the law of reflection basically defines light reflections phenomena.

ziro4ka [17]3 years ago
4 0

light bounces off a boundary and light direction is determined using the lawof reflection

I hope it helped

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In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he
lana [24]

Answer:8.28 km

Explanation:  

Given

First it drifts 45^{\circ} 2.5 km

r_1=2.5cos45 i+2.5sin45 j

Secondly it drifts 60^{\circ} 4.70 km

r_{12}=4.7cos60 i-4.7sin60 j

After that it drifted along east direction 5.1 km

r_{23}=5.1 i

After that it drifts 55^{\circ} 7.2 km

r_{34}=-7.2cos55 i-7.2sin55 j

After that it drifts 5^{\circ} 2.8 km

r_{54}=-2.8cos5 i+2.8sin5 j

r_{5O}=\left [ 2.5cos45+4.7cos60+5.1-7.2cos55-2.8cos5\right ]\hat{i}+\left [ 2.5sin45-4.7sin60-7.2sin55+2.8sin5\right ]\hat{j}

r_{5O}=2.299\hat{i}-7.95\hat{j}

|r_{5O}|=8.28 km

for direction

tan\theta =\frac{7.95}{2.299}=3.4580

\theta =73.87^{\circ} south of east

7 0
3 years ago
Read 2 more answers
A 7-hp (shaft) pump is used to raise water to an elevation of 19 m. if the mechanical efficiency of the pump is 82 percent, dete
Hoochie [10]
Convert the units of power,W = 7 hp = 7 * 745.69 = 5219.83 WCalculate the power input to the pump using the efficiency of the pump equationn=Wpump/Wshaft
Substitute 0.82 for n and 5219.83 for Wshaft0.82=Wpump/5219.83   Wpump=0.82*5219.83=4280.26 WCalculate the mass flow ratein=Wpump/(gz_2 )Where g is the acceleration due to gravity, and z_2 is the elevation of water. Substitute 4280.26 for Wpump, 9.81 m/s^2 for g, and 19m for z_2in = 4280.26 / 9.81 * 19 = 22.9640 m^3/sCalculate the volume flow rate of waterV=m/ρWhere ρ is the density of water. Substitute 22.9640  m^3/s for in and 1000  m^3/kg for ρ, we get V = 22.9640 / 1000 = 0.0230 kg/sTherefore, the volume flow rate of water is 0.0230 kg/s
7 0
3 years ago
Please help me
bonufazy [111]

Переходи на сайте irkmix.top и получай много эмоций из Russia/

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3 years ago
True or False<br><br> The greater the speed of an object, the less kinetic energy it possesses.
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That is true because if the object is moving at Forceful speeds than it will lose more of its kinetic energy
3 0
3 years ago
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Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init
Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ=(m+\frac{1}{2})\frac{lamda}{n}  ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= \frac{3*lamda}{2}

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= \sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }

δ= \sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}

δ= \sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }

δ= \sqrt{4900+43^2}-\sqrt{4900+23^2}

δ= \sqrt{4900+1849}-\sqrt{4900+529}

δ= \sqrt{6749}-\sqrt{5429}

δ=  82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= \frac{3*lamda}{2}

δ= 8.47

8.47 = \frac{3*lamda}{2}

λ = \frac{8.47*2}{3}

λ = 5.65m

3 0
3 years ago
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