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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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All matter has electrical and magnetic properties because the atoms that make up matter are held together by electromagnetic for
leva [86]

Answer:

Charge

Explanation:

Charge is a fundamental property of all matter. All matter has electrical and magnetic properties because the atoms that make up matter are held together by electromagnetic forces.

These charges are usually positive and negative charges. When these charges which make up an atom (positive and negative) are equal, the atom is said to be electrically neutral. When positive charge is greater than negative charge, the atom is said to positively charged. Also, if the number of negative charges are more in an atom, the atom is said to be negatively charged.

7 0
3 years ago
The experiments Galileo performed, such as rolling a ball down an inclined plane, are important because they
anyanavicka [17]

Galileo Galilei is one of the key figures in the history of Science, being the first to apply the experimental-mathematical scientific method. He carried out experiments and careful observations in kinematics (his studies on the trajectory of projectiles are famous) and dynamics (it should be noted his careful experiments with inclined planes), establishing the first law of Dynamics (which Newton will later collect and refine in his Principles); and in Astronomy, with which he could unequivocally support the heliocentric theory.

His experiments were addressed by methodologies that allowed him to precisely find his mathematical calculations and to verify theories he was developing over time. His manuscripts were key to disseminate the applied method and extrapolate them to other scientific areas.

Therefore the correct answer is C.

7 0
3 years ago
An object undergoes two successive displacements:
Gre4nikov [31]

The magnitude of the net displacement is 95.3 m

Explanation:

To find the magnitude of the net displacement, we have to resolve each of the two displacements into the horizontal and vertical direction first.

1st displacement is:

d_1=79 m at 16.9^{\circ}

So its components are

d_{1x}=(79)(cos 16.9^{\circ})=75.6 m\\d_{1y}=(79)(sin 16.9^{\circ})=23.0 m

2nd displacement is:

d_2=16.7 m at 31.1^{\circ}

So its components are

d_{2x}=(16.7)(cos 31.1^{\circ})=14.3 m\\d_{2y}=(16.7)(sin 31.1^{\circ})=8.6 m

Therefore, the x- and y-components of the net displacement are:

d_x=d_{1x}+d_{2x}=75.6+14.3=89.9 m\\d_y=d_{1y}+d_{2y}=23.0+8.6=31.6 m

Therefore, the magnitude of the final displacement is:

d=\sqrt{d_x^2+d_y^2}=\sqrt{(89.9)^2+(31.6)^2}=95.3 m

Learn more about displacement:

brainly.com/question/3969582

#LearnwithBrainly

8 0
3 years ago
A light-rail commuter train accelerates at a rate of 1.35 m/s. D A 33% Part (a) How long does it take to reach its top speed of
Dennis_Churaev [7]

Answer:

a) 17.49 seconds

b) 13.12 seconds

c) 2.99 m/s²

Explanation:

a) Acceleration = a = 1.35 m/s²

Final velocity = v = 85 km/h = 85\frac{1000}{3600}=23.61\ m/s

Initial velocity = u = 0

Equation of motion

v=u+at\\\Rightarrow 23.61=0+1.35t\\\Rightarrow t=\frac{23.61}{1.35}=17.49\ s

Time taken to accelerate to top speed is 17.49 seconds.

b) Acceleration = a = -1.8 m/s²

Initial velocity = u = 23.61\ m/s

Final velocity = v = 0

v=u+at\\\Rightarrow 0=23.61-1.8t\\\Rightarrow t=\frac{23.61}{1.8}=13.12\ s

Time taken to stop the train from top speed is 13.12 seconds

c) Initial velocity = u = 23.61 m/s

Time taken = t = 7.9 s

Final velocity = v = 0

v=u+at\\\Rightarrow 0=23.61+a7.9\\\Rightarrow a=\frac{-23.61}{7.9}=-2.99\ m/s^2

Emergency acceleration is 2.99 m/s² (magnitude)

6 0
3 years ago
Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm
Burka [1]

Answer:

The atmospheric pressure is 0.97622\times10^{5}\ Pa.

Explanation:

Given that,

Atmospheric pressure P_{atm}= 1.013\times10^{5}\ Pa    

drop height h'= 27.1 mm

Density of mercury \rho= 13.59 g/cm^3

We need to calculate the height

Using formula of pressure

p = \rho g h

Put the value into the formula

1.013\times10^{5}=13.59\times10^{3}\times9.8\times h

h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}

h=0.76\ m

We need to calculate the new height

h''=h - h'

h''=0.76-27.1\times10^{-3}

h''=0.76-0.027

h''=0.733\ m

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

P=\rho g h

Put the value into the formula

P= 13.59\times10^{3}\times9.8\times0.733

P=0.97622\times10^{5}\ Pa

Hence, The atmospheric pressure is 0.97622\times10^{5}\ Pa.

7 0
3 years ago
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