Ask question

Ask question

All categories

3 years ago

15

The difference in energy () between vibrational energy levels is determined by the nature of the bond. If a photon of light poss

esses exactly this amount of energy, the bond can absorb the photon to promote a excitation. True or false?

1 answer:

forsale [732]3 years ago

5 0

Answer:

the statement is true

Explanation:

The bonds in the molecules give rise to irrational energy levels, are quantized, that is, they are discrete

In order for a photon to create a transition between the given levels it has to have the same energy as these two levels, it is in this case if the photon has the energy of the bond can be promised to the electron and be absorbed.

In summary the statement is true

You might be interested in

1. Which object is farthest from the origin at t=2sec

Stolb23 [73]

Answer:

that one i know only pe not that sorry again

6 0

2 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

3 0

2 years ago

You drive a car 640 m to the east, then 340 m to the north what is the magnitude of your displacement

mixer [17]

Magnitude of displacement = $\sqrt{640^2 + 340^2}$

Adding the squares gives displacement = $\sqrt{525,200}$

Displacement = $\sqrt{525,200}$ ≈ 724.7m

6 0

2 years ago

according to isaac newtons theory of gravitation the force exerted between two objects is dependent on- A- an objects weight. B-

puteri [66]

A an objects weight hope this helps! :D

3 0

2 years ago

What is the speed of an object after falling (from rest) a distance of 9.80 meters near the surface of the Earth? Assume air

madam [21]

Hi there!

We can use the kinematic equation:

$v_f^2 = v_i^2 + 2ad$

vf = Final velocity (? m/s)

vi = initial velocity (0 m/s, dropped from rest)

a = acceleration (due to gravity, 9.8 m/s²)

d = distance (9.8 m)

Simplify the equation to solve for vf:

$v_f^2 = 0 + 2ad\\\\v_f = \sqrt{2ad}$

Substitute in the given values:

$v_f = \sqrt{2(9.8)(9.8)} = \boxed{13.86 m/s}$

8 0

2 years ago