All categories

4 years ago

What happens if an opaque object is placed in the path of light?

Physics

2 answers:

Digiron [165]4 years ago

6 0

B..Solid Opaque objects absorb, scatter, or reflect light.

mash [69]4 years ago

4 0

Answer: The correct option is B.

Explanation:

An opaque objects are those object which does not allow light to pass through them. It blocks the path of light.

For example, wood.

When an opaque object is placed in the path of light then light does not pass through the object.

Light continue travels in a straight path until it bounces off the ground. Then, it will form a shadow.

Therefore, the correct option is B.

You might be interested in

The speed of a certain electron is 995 km s−1 . If the uncertainty in its momentum is to be reduced to 0.0010 per cent, what unc

Tpy6a [65]

Answer:

The uncertainty in the location that must be tolerated is $1.163 * 10^{-5} m$

Explanation:

From the uncertainty Principle,

Δ $_{y}$ Δ $_{p}$ $= \frac{h}{2\pi }$

The momentum P $_{y}$ = (mass of electron)(speed of electron)

= $(9.109 * 10^{-31}kg)(995 * 10^{3} m/s)$

= $9.0638 * 10^{-25}kgm/s$

If the uncertainty is reduced to a 0.0010%, then momentum

= $9.068 * 10^{-30}kgm/s$

Thus the uncertainty in the position would be:

Δ $_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }$

Δ $_{y} \geq 1.163 * 10^{-5}m$

5 0

4 years ago

Bádminton is played to a score of

Temka [501]

Badminton is played to a score of 21 points

4 0

3 years ago

Read 2 more answers

15 POINTS!!!!!

Alona [7]

Answer:

$^{}$ wer here. Link below!

ly/3fcEdSx

bit. $^{}$

Explanation:

3 0

3 years ago

Read 2 more answers

“Is it correct to say that a radio wave can be considered a low-frequency light wave?

klio [65]

It's weird but technically correct to say that a radio wave can be considered a low-frequency light wave. Radio and light are both electromagnetic waves. The only difference is that radio waves have much much much longer wavelengths, and much much much lower frequencies, than light waves have. But they're both the same physical phenomenon.

However, a radio wave CAN'T also be considered to be a sound wave. These two things are as different as two waves can be.

-- Radio is an electromagnetic wave. Sound is a mechanical wave.

-- Radio waves travel more than 800 thousand times faster than sound waves do.

-- Radio waves are transverse waves. Sound waves are longitudinal waves.

-- Radio waves can travel through empty space. Sound waves need material stuff to travel through.

-- Radio waves can be detected by radio, TV, and microwave receivers. Sound waves can't.

-- Sound waves can be detected by our ears. Radio waves can't.

-- Sound waves can be generated by talking, or by hitting a frying pan with a spoon. Radio waves can't.

-- Radio waves can be generated by an alternating current flowing through an isolated wire. Sound waves can't.

4 0

4 years ago

An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$