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1 year ago

What is double-slit experiment?

1 answer:

4 0

The double-slit experiment shows that both matter and light can exhibit properties of conventionally defined waves and particles.

The double-slit experiment is a part of a class of "double path" experiments in which a wave is split into two separate waves that later combine to form a single wave (the wave is typically composed of many photons and is better known as a wave front, which should not be confused with the wave properties of the individual photon).

Isaac Newton's corpuscular theory of light, which had previously prevailed as the accepted explanation of light transmission in the 17th and 18th centuries, was defeated by double-slit experiment , which was conducted in the early 1800s.

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Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

2 years ago

What is power?

Blizzard [7]

Answer:

Explanation:

P=Work/Time

The rate at which work is done matches that.

6 0

3 years ago

Can somebody please help?

DedPeter [7]

Answer:

Im not sure

Explanation:

I don't take physics cuz im in 9th grade. so. idk but I will find out and come back with an answer.

7 0

2 years ago

You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sin

faust18 [17]

Answer:

Frequency required will be 2421.127 kHz

Explanation:

We have given inductance $L=0.450H=0.45\times 10^{-3}H$

Current in the inductor $i=1.90mA=1.90\times 10^{-3}A$

Voltage v = 13 volt

Inductive reactance of the circuit $X_l=\frac{v}{i}$

$X_l=\frac{13}{1.9\times 10^{-3}}=6842.10ohm$

We know that

$X_l=\omega L=2\pi fL$

$2\times 3.14\times f\times 0.45\times 10^{-3}=6842.10$

f = 2421.127 kHz

6 0

3 years ago

In a collision between two unequal masses, which mass receives a greater magnitude impulse?

Contact [7]

<span>The answer to this question depends upon Newton's third law of motion. For every action, there's an equal and opposite reaction. Because of this law, during the collision between two unequal masses, the impulse that each mass receives will be of equal magnitude and and opposite sign.</span>

8 0

2 years ago