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2 years ago

Which phenomenon can only be explained by assuming that light is quantized

Physics

1 answer:

Serjik [45]2 years ago

3 0

<h2>Answer: The Photoelectric Effect</h2>

Light can be considered as a wave or as particles, in this context Einstein proposed that light behaves like a stream of particles called photons with an energy, in order to correctly explain the photoelectric effect.

This fenomenom consists in the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.

So, if we consider light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.

This means the photoelectric effect can only be explained based on the corpuscular model of light, that is, light is quantized.

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An electrical short cuts off all power to a submersible diving vehicle when it is a distance of 28 m below the surface of the oc

skelet666 [1.2K]

Answer:

F=126339.5N

Explanation:

to find the necessary force to escape we must make a free-body diagram on the hatch, taking into account that we will match the forces that go down with those that go up, taking into account the above we propose the following equation,

Fw=W+Fi+F

where

Fw= force or weight produced by the water column above the submarine.

to fint Fw we can use the following ecuation

Fw=h. γ. A

h=distance

γ= specific weight for seawater = 10074N / m ^ 3

A=Area

Fw=28x10074x0.7=197467N

w is the weight of the hatch = 200N

Fi is the internal force of the submarine produced by the pressure = 1atm = 101325Pa for this we can use the following formula

Fi=PA=101325x0.7=70927.5N

finally the force that is needed to open the hatch is given by the initial equation

Fw=W+Fi+F

F=Fw-W+Fi

F=197467N-200N-70927.5N

F=126339.5N

6 0

2 years ago

A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington

Roman55 [17]

Let $\theta$ be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

$\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

The current has velocity vector (relative to the Earth)

$\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath$

The swimmer's resultant velocity (her velocity relative to the Earth) is then

$\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}$

$\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath$

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

$\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ$

which is approximately 41º west of north.

6 0

2 years ago

A baggage handler throws a 15 kg suitcase along the floor of an airplane luggage compartment with a speed of 1.2 m/s. The suitca

Hatshy [7]

Answer:

0.0367

Explanation:

The loss in kinetic energy results into work done by friction.

Since kinetic energy is given by

KE=0.5mv^{2}

Work done by friction is given as

W= umgd

Where m is the mass of suitacase, v is velocity of the suitcase, g is acceleration due to gravity, d is perpendicular distance where force is applied and u is coefficient of kinetic friction.

Making u the subject of the formula then we deduce that

$u=\frac {v^{2}}{2gd}$

Substituting v with 1.2 m/s, d with 2m and taking g as 9.81 m/s2 then

$u=\frac {1.2^{2}}{2*9.81*2}=0.0366972477064\approx 0.0367$

Therefore, the coefficient of kinetic friction is approximately 0.0367

7 0

2 years ago

Determine the force P required to maintain the 200-kg engine in the position for which θ = 30°. The diameter of the pulley at B

gregori [183]

Answer:

The force P required is 1759.22 N

Explanation:

The missing diagram is seen in the first image below.

From the second image, we can see the schematic diagram of the engine hanging over the pulley.

To start with determining the value of the angle ∝;

$tan \ \alpha = \dfrac{CD}{BD}$

where;

BD = AB-AD

Then;

$tan \ \alpha = \dfrac{CD}{AB-AD}$

$\alpha = tan^{-1} \bigg(\dfrac{CD}{AB-AD} \bigg )$

replacing their respective values, where;

CD = 2 sin 30° m, AB = 2m and AD = 2 cos 30° m

$\alpha = tan^{-1} \bigg(\dfrac{2 \ sin \ 30^0}{2-2 \ cos \ 30^0} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{2-1.732} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{0.268} \bigg )$

$\alpha = tan^{-1} \bigg(3.73\bigg )$

$\alpha \simeq 75^0$

From the third diagram attached below:

The tension occurring in the thread BC is equal to force P

$T_{BC} = P$

Using the force equilibrium expression along the horizontal direction.

$\sum F_x = 0\\\\ -T_{AC} \ cos \ 30^0 + Pcos \alpha = 0$

replacing the value of $\alpha \simeq 75^0$

$-T_{AC} \ cos 30^0 + P cos 75^0 = 0$

$P \ cos \ 75^0 = T_{AC} \ cos \ 30^0$

$P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0} \ \ \ - - - (1)$

Along the vertical direction, the force equilibrium equation can be expressed as:

$\sum F_y =0$

$-W + P \ sin \alpha + T_{AC} \ sin \ 30^0 = 0$

$W = P \ sin \ \alpha + T_{AC} \ sin \ 30^0$

replacing $\alpha \simeq 75^0$ and $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$W =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

Also, replacing W for (200 × 9.81) N

$200 \times 9.81 =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

$200 \times 9.81 = T_{AC} \ cos \ 30^0 \ tan \ 75^0 + T_{AC} \ sin \ 30^0$

$1962= T_{AC} \ ( cos \ 30^0 \ tan \ 75^0 + \ sin \ 30^0)$

$1962= T_{AC} \ (0.8660\times 3.732 + 0.5)$

$1962= T_{AC} \ (3.231912 + 0.5)$

$1962= T_{AC} \ (3.731912)$

$T_{AC} = \dfrac{1962}{ \ (3.731912)}$

$T_{AC} = 525.736 \ N$

From $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \times0.866}{0.2588}$

P = 1759.22 N

Thus, the force P required is 1759.22 N

6 0

1 year ago

A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is

aivan3 [116]

A. a = 1.3 m/s^2; FN = 63.1 N

4 0

2 years ago

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