The theory that light exists in quantized "chunks" was proposed in response

Relationship between

vlabodo [156]

Answer:

The energy of a wave is proportional to the square of its amplitude

Explanation:

The graph in this problem is missing.

However, we can describe what is the relationship between amplitude and energy of a wave.

In fact, for a wave, the energy carried by it is proportional to the square of its amplitude:

$E\propto A^2$

where

E is the energy of the wave

A is the amplitude

This means that, for instance:

- as the amplitude of the wave increases by a factor of 2, the energy carried by the wave increases by a factor of 4

- As the amplitude of the wave increases by a factor of 3, the energy carried by the wave increases by a factor of 9

... and so on.

5 0

3 years ago

Read 2 more answers

Compare and contrast the electric force with the gravitational force

Olenka [21]

Similarities:

- The gravitational force and the electric force have a similar form.

In fact, gravitational force is given by:

$F=G\frac{m_1 m_2}{r^2}$

where G is the gravitational constant, m1 and m2 are the masses of the two objects, r is the separation between the two objects.

The electric force is given by:

$F=k\frac{q_1 q_2}{r^2}$

where k is the Coulomb constant, q1 and q2 are the charges of the two objects, r is the separation between the two objects.

As we see, the two forces have same dependency on the distance (inversely proportional to the square of the distance)

- Both forces are non-contact forces: this means that in both cases, the two objects do not to be in contact with each other, because the force is still acting from a distance also.

Differencies:

- The gravitational force is attractive only; on the contrary, the electric force can be either attractive or repulsive, depending on the relative sign of the two charges. In fact, when the two charges have same sign, the force is repulsive; when the two charges have opposite sign, the force is attractive.

- The gravitational force is much weaker than the electric force. This can be seen by comparing the value of the constant in the two formulas; we have:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ for the gravitational force

$k=8.99 \cdot 10^9 Nm^2C^{-2}$ for the electric force

As we see, the electric force is much stronger than the gravitational force.

6 0

3 years ago

You will often have to pay a small fee<br> to obtain a web address.<br> True<br> False

CaHeK987 [17]

Answer:

true?

Explanation:

4 0

3 years ago

Consider the transition from the energy levels n = 3 to n = 6. what is the wavelength associated with this transition, in nm?

Anuta_ua [19.1K]

In quantum chemistry, when the energy of electron changes by moving from orbital to orbital, it emits or absorbs energy in the form of light. When this happens, you can measure its wavelength. Suppose an atom is heated. This excites the electron so it moves up higher to the next orbital. As a result, it would emit light. When its energy lowers and moves one orbital lower, it would absorb light. The equation to determine the wavelength is called the Rydberg formula.

1/wavelength = R(1/n1^2 - 1/n2^2), where
R is an empirical constant called Rydberg's constant equal to 1.9074 × 10^7 m^-1
n1 and n2 are orbitals of the atoms such that n2 is always greater than n1.

Substituting the values:

1/wavelength = 1.9074 × 10^7 (1/3^2 - 1/6^2)
wavelength = 1.09 × 10^-6 m or 1.09 micrometers

3 0

3 years ago

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.310 m and then accelerates through this dista

Nastasia [14]

Answer:A)u =4.295m/s , B)a = 29.746m/s² C) F=3,153N

Explanation:

Using the kinematic expression

v² = u² - 2as

where

u = initial velocity

v = final velocity

s = distance

g = acceleration due to gravity .

Given that he reaches a height of 0.940 m above the floor,

the final velocity = 0

Here, acceleration due to gravity is acting in opposite the initial direction of motion. So, a=-9.81 m/s.

v² = u² + 2as

0² - u² = 2 (- 9.81) × 0.940

- u² = 2 × - 9.81 × 0.920

- u² = -18.4428

cancelling the minus in both sides , we have that

u² = 18.4428

u = √18.4428

u =4.295m/s

(b) His acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2

Using v² = u² + 2as

where u = initial speed of basketball player before lengthening = 0 m/s,

v = final speed of basketball player after lengthening = 4.295m/s,

a = acceleration while straightening his legs

s = distance moved during lengthening = 0.310m

v² = u² + 2as

a = (v² - u²)/2s

a = (4.29m/s)² - (0 m/s)²)/(2 × 0.310m)

a = (18.4428 m²/s² - 0 m²/s²)/(0.62 m)

a = (18.4428 m²/s²/(0.62 m)

a = 29.746m/s²

c) The force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N

Force= mass x acceleration.

F = 106 kg X 29.746m/s²

F = 3,153.076 rounded to 3,153N

8 0

3 years ago