One us of alkali metals

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

4 0

4 years ago

A helium atom (mass 4.0 u) moving at 598 m/s to the right collides with an oxygen molecule (mass 32 u) moving in the same direct

labwork [276]

Answer:

Speed of the helium after collision = 246 m/s

Explanation:

Given that

Mass of helium ,m₁ = 4 u

u₁=598 m/s

Mass of oxygen ,m₂ = 32 u

u₂ = 401 m/s

v₂ =445 m/s

Given that initially both are moving in the same direction and lets take they are moving in the right direction.

Speed of the helium after collision = v₁

There is no any external force on the masses that is why the linear momentum will be conserve.

Initial linear momentum = Final linear momentum

P = m v

m₁u₁+m₂u₂ = m₁v₁+m₂v₂

598 x 4 + 32 x 401 = 4 x v₁+ 32 x 445

v₁ = 246 m/s

Speed of the helium after collision = 246 m/s

6 0

3 years ago

Sound waves are mechanical waves in which the particles in the medium vibrate in a direction parallel to the direction of energy

ch4aika [34]

C longitudnal waves

4 0

3 years ago

Read 2 more answers

A beam of monochromatic light is aimed at a slit of width and forms a diffraction pattern. In which case is the width of the cen

AysviL [449]

When the incident light is yellow the width of the central band greater. Single-wavelength light sources are known as monochromatic lights, where mono stands for one and chroma for color. Monochromatic lights are defined as visible light that falls inside a specific range of wavelengths. It has a wavelength that falls within a constrained wavelength range.

A laser beam is the ideal illustration of monochromatic light. A monochromatic light beam produced by a single atomic transition with a particular single wavelength is what makes up a laser. A color scheme that consists solely of different shades of one color is referred to as monochromatic.

To learn more about monochromatic, click here.

brainly.com/question/23624834

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4 0

2 years ago

You can think of the work-kinetic energy theorem as the second theory of motion, parallel to Newton's laws in describing how out

kiruha [24]

Answer:

a) 4 289.8 J

b) 4 289.8 J

c) 6 620.1 N

d) 411 186.3 m/s^2

e) 6 620.1 N

Explanation:

Hi:

a)

The kinetic energy of the bullet is given by the following formula:

K = (1/2) m * v^2

With

m = 16.1 g = 1.61 x 10^-2 kg

v = 730 m/s

K = 4 289.8 J

b)

the work-kinetic energy theorem states that the work done on a system is the same as the differnce in kinetic energy of the same. Since the initial state of the bullet was at zero velocity (it was at rest) Ki = 0, therefore:

W = ΔK = Kf - Ki = 4 289.8 J

c)

The work done by a force is given by the line intergarl of the force along the trayectory of the system (in this case the bullet).

If we consider a constant force (and average net force) directed along the trayectory of the bullet, the work and the force will be realted by:

W = F * L

Where F is the net force and L is the length of the barrel, that is:

F = (4 289.8 J) / (64.8 cm) = (4 289.8 Nm) / (0.648 m) = 6620.1 N

d)

The acceleration can be found dividing the force by the mass:

a = F/m = (6620.1 N) /(16.1 g) = 411 186.3 m/s^2

e)

The force will have a magnitude equal to c) and direction along the barrel towards the exit

5 0

3 years ago