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2 years 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.

To know more about double-slit experiment follow the link:

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A 3.0-kg block is on a horizontal surface. The block is at rest when, at t = 0, a force (magnitude P = 12 N) acting parallel to

GarryVolchara [31]

Explanation:

It is given that,

Mass of the block, m = 3 kg

Initially, the block is at rest, u = 0

Force acting on the block, P = 12 N

The coefficient of kinetic friction between the block and the surface is, $\mu_k=0.2$

We need to find the rate is the force P doing work on the block at t = 2.0 s. The rate at which work is done is called the power. Let is equal to P'

Frictional force acting on the block, $f=\mu_k mg$

$f=0.2\times 3\ kg\times 9.8\ m/s^2=5.88\ N$

So, the net force acting on the block, F = P - f

$F=12-5.88=6.12\ N$

Let a is the acceleration of the block, $a=\dfrac{F}{m}$

$a=\dfrac{6.12}{3}=2.04\ m/s^2$

Let v is the velocity of the block after 2 seconds. So,

$v=u+at$

$v=0+2.04\times 2$

v = 4.08 m/s

Power, $P'=\dfrac{W}{t}=\dfrac{F.d}{t}=F.v$

$P'=12\ N\times 4.08\ m/s=48.96\ Watts$

So, the force P is doing work on the block at the rate of 48.96 watts.

3 0

3 years ago

The absolute magnitude of a star________.

serg [7]

<span>The absolute magnitude of a star is how bright it would appear to us
if it were located ten parsecs (about 32.6 light years) from us. So it's
a way of treating all stars equally ... on a "level playing field" ... and it
describes each star's actual brightness. </span>

8 0

3 years ago

A hydrogen atom contains a single electron that moves in a circular orbit about a single proton. Assume the proton is stationary

riadik2000 [5.3K]

Answer:

Radius between electron and proton $= 6.804\times 10^{-10}m$

Explanation:

The motion of the electron is carried out in the orbit due to the balancing of the electrostatic force between the proton and the electron and the centripetal force acting on the electron.

The electrostatic force is given as = $\frac{kq_1q_2}{r^2}$

Where,

k = coulomb's law constant (9×10⁹ N-m²/C²)

q₁ and q₂ = charges = 1.6 × 10⁻¹⁹ C

r = radius between the proton and the electron

Also,

Centripetal force on the moving electron is given as:

= $\frac{m_eV^2}{r}$

where,

$m_e$ = mass of the electron (9.1 ×10⁻³¹ kg)

V = velocity of the moving electron (given: 6.1 ×10⁵ m/s)

Now equating both the formulas, we have

$\frac{kq_1q_2}{r^2}$ = $\frac{m_eV^2}{r}$

⇒ $r = \frac{kq_1q_2}{m_eV^2}$

substituting the values in the above equation we get,

$r = \frac{9\times 10^{9}\times (1.6\times 10^{-19})^2}{9.1\times 10^{-31}\times (6.1\times 10^5)^2}$

⇒ $r = 6.804\times 10^{-10}m$

3 0

3 years ago

If you calculate W, the amount of work it took to assemble this charge configuration if the point charges were initially infinit

sp2606 [1]

Answer:

W = 0×(kq2L)

Explanation:

We know that the work to assemble a charge configuration of two charges a distance r from each other is simply W = kq2/r

If we want to assemble three charges A, B, and C. It's necessary to consider the distances between them

WABC = kq2/(rAB + rAC + rBC)

So, to assemble four charges A, B, C, & D....

WABCD = kq2/(rAB + rAC + rAD + rBC + rBD + rCD)

Considering a square charge configuration with sides L, such as in figure attached A, B, & C are positive & D is negative

rAB = L

rAC = L√2

rAD = L (-)

rBC = L

rBD = L√2 (-)

rCD = L (-)

⇒ W = kq2/(L + L√2 + (-L) + L + (-L√2) + (-L)

⇒ ∴ W = 0 × (kq2/L)

This way, working through each option...

(a)

The positive charges are equidistant from each other at a distance of L.

rAB = L

rAC = L

rAD = ½L⋅sin(60) (-)

rBC = L

rBD = ½L⋅sin(60) (-)

rCD = ½L⋅sin(60) (-)

Wa = kq2/(3L - (3/2)L⋅(0.866))

⇒ ∴ Wa = (1/1.7) × (kq2/L) = (0.5879)× (kq2/L)

(b)

rAB = L

rAC = 2L

rAD = 3L (-)

rBC = L

rBD = 2L (-)

rCD = L (-)

Wb = kq2/(4L - 6L)

⇒ ∴ Wb = (-1/2) × (kq2/L) = (-0.5)× (kq2/L)

(c)

The factor doesn't matter, so Wc = 0 × (kq2/L)

In this case, the greater work is actually the less work. Therefore, the positive work represents the amount of work the system actually exhibits, that we don't have to do. If there is negative work, we have to make up that work in order to place the charges as desired.

This way, charge configuration (a) requires the least amount of work.

5 0

3 years ago

Describing motion with velocity-time graphs.

irina1246 [14]

This link should help you out!!

http://mszopiak.weebly.com/uploads/5/5/8/1/55815541/p_vtgraphskey_mszopiak.pdf

5 0

3 years ago