Ask question

Ask question

All categories

2 years ago

5

That light has a dual nature is referring to light: having energy and momentum. having high- or low-energy photons. acting as wa

ves and particles. undergoing pair production.

1 answer:

zloy xaker [14]2 years ago

5 0

Answer:

Option C, acting as waves and particles

Explanation:

Light has dual nature because it acts both as a wave and particle. It has high energy particle i.e photons and it also behave as an electromagnetic wave. This property of light is studied under the quantum mechanics. Einstein also proved that light is a stream of photons possessing both electrical and magnetic properties.

Hence, option C is correct

You might be interested in

A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is

AveGali [126]

Explanation:

For this problem, use the first law of thermodynamics. The change in energy equals the increase in heat energy minus the work done.

ΔU=Q−W

We are not given a value for work, but we can solve for it using the force and distance. Work is the product of force and displacement.

W=FΔx

W=3N×2m

W=6J

Now that we have the value of work done and the value for heat added, we can solve for the total change in energy.

ΔU=Q−W

ΔU=10J−6J

ΔU=4J

Answer is 4J

i think this may help you very much

3 0

2 years ago

Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain

Luden [163]

Answer:

$P=1362\ W$

$t'=251.659\ s$ is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, $T_i=18^{\circ}C$

time taken to vapourize half a liter of water, $t=18\ min=1080\ s$

desity of water, $\rho=1\ kg.L^{-1}$

So, the givne mass of water, $m=1\ kg$

enthalpy of vaporization of water, $h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}$

specific heat of water, $c=4180\ J.kg^{-1}.K^{-1}$

Amount of heat required to raise the temperature of given water mass to 100°C:

$Q_s=m.c.\Delta T$

$Q_s=1\times 4180\times (100-18)$

$Q_s=342760\ J$

Now the amount of heat required to vaporize 0.5 kg of water:

$Q_v=m'\times h_{fg}$

where:

$m'=0.5\ kg=$ mass of water vaporized due to boiling

$Q_v=0.5\times 2256.4$

$Q_v=1.1282\times 10^{6}\ J$

Now the power rating of the boiler:

$P=\frac{Q_s+Q_v}{t}$

$P=\frac{342760+1128200}{1080}$

$P=1362\ W$

Now the time required to heat to boiling point form initial temperature:

$t'=\frac{Q_s}{P}$

$t'=\frac{342760}{1362}$

$t'=251.659\ s$

6 0

3 years ago

777dan777 [17]

<span>One leg is = 12 m, and the other leg is 16 m. </span>

3 0

2 years ago

A car moves at a constant speed of 90km/h from a starting point. Another car moves at 70km/h after 2hours from the same starting

emmainna [20.7K]

Answer:

400

Explanation:

5 0

2 years ago

g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr

Lorico [155]

To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

$a = \frac{GM}{r^2}$

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

$\frac{da}{dr} = -2\frac{GM}{r^3}$

$da = -2\frac{GM}{r^3}dr$

At the same time since Newton's second law we know that:

$F_w = ma$

Where,

m = mass

a =Acceleration

From the previous value given for acceleration we have to

$F_W = m (\frac{GM}{r^2} ) = 600N$

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

$dF_W = mda$

$dF_W = m(-2\frac{GM}{r^3}dr)$

$dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})$

$dF_W = -2F_W(\frac{dr}{r})$

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

$dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})$

$dF_W = -0.3N$

Therefore there is a weight loss of 0.3N every kilometer.

4 0

3 years ago