SOME HELP ME WITH MY QUIZ

20. GP A sinusoidal wave traveling in the negative x direction (to the left) has an amplitude of 20.0cm, a wavelength of 35.0cm,

umka21 [38]

The angular frequency of the wave is determined as 75.4 rad/s.

<h3>What is wave function?</h3>

A wave function is a mathematical equation for the motion of the wave.

y(x, t) = A sin(kx + ωt + Φ)

where;

ω is angular speed
k is angular wavenumber
Φ is phase angle

<h3>What is angular frequency?</h3>

The angular frequency is the angular displacement of any wave element per unit of time or the rate of change of the waveform phase.

<h3>Angular frequency</h3>

ω = 2πf

ω = 2π(12)

ω = 75.4 rad/s

Thus, the angular frequency of the wave is determined as 75.4 rad/s.

Learn more about angular frequency here: brainly.com/question/3654452

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7 0

1 year ago

A 2.13-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N

ANTONII [103]

Answer:

17.54N in -x direction.

Explanation:

Amplitude (A) = 3.54m

Force constant (k) = 5N/m

Mass (m) = 2.13kg

Angular frequency ω = √(k/m)

ω = √(5/2.13)

ω = 1.53 rad/s

The force acting on the object F(t) = ?

F(t) = -mAω²cos(ωt)

F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)

F(t) = -17.65 * cos (5.355)

F(t) = -17.57N

The force is 17.57 in -x direction

5 0

3 years ago

Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at

PSYCHO15rus [73]

Answer:

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

$\mu= \frac{m_1m_2}{m_1+m_2}$

now, since both the masses have mass m

therefore,

$\mu= \frac{m^2}{2m}$

= m/2

The final K.E. of the particles is

$KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2$

Distance between two particles is d and the gravitational potential energy between them is given by

$PE_{Gravitational}= \frac{Gmm}{d}$

By law of conservation of energy we have

$KE_{initial}+KE_{final}= PE_{gravitaional}$

Now plugging the values we get

$0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}$

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

$=\sqrt{\frac{Gm}{d} }$

This the required relation between G,m and d

5 0

2 years ago

From the window of a house that is placed 15 m

kow [346]

Answer:

a) 52.915 m

b) The vertical velocity is approximately 21.092 m/s

The resultant velocity is approximately 26.5 m/s

Explanation:

a) The height of the window in the house from which the water was thrown = 15 m

The speed of the stream of water thrown = 20 m/s

The angle at which the water was thrown = 37° over the horizontal

The acceleration due to gravity, g = 10 m/s²

a) The distance from the base of the house at which the water will fall is given as follows;

y = y₀ + u·t·sin(θ) + 1/2·g·t²

Where;

y = The vertical height reached

u = The initial velocity

t = Time of flight

From the point the steam of water is thrown, we get;

y₀ = 15 m

Therefore;

y = 15 + 20 × t × sin(37°) - 1/2 × 10 × t²

y = 15 + 20 × t × sin(37°) - 5 × t²

When y = 0, Ground level, we get

0 = 15 + 20 × t × sin(37°) - 5 × t²

5·t² - 20×sin(37°)×t -15 = 0

∴ t = (20 ×sin(37°) ± √((-20 × ·sin(37°))² - 4 × (5) × (-15)))/(2 × 5)

t ≈ 3.3128302, or t ≈ 0.906

Therefore, the time of flight of the water, t ≈ 3.3128302 seconds

The distance from the base of the house at which the water will fall = The horizontal distance travelled by the water, x

x = u·cos(θ)×t

∴ x = 20 × cos(37°) × 3.3128302 ≈ 52.915

The distance from the base of the house at which the water will fall = x ≈ 52.915 m

b) The velocity at which the water will reach the ground, 'v', is given as follows;

The vertical velocity, $v_y$ = u·sin(θ)·t - g·t

At the ground, t ≈ 3.3128302 seconds

∴ $v_y$ = 20 × sin(37) - 10 × 3.3128302 ≈ -21.092

The vertical velocity at which the water will reach the ground, $v_y$ ≈ 21.092 m/s (downwards)

The resultant velocity, v = √( $v_y$ ² + vₓ²)

∴ v = √(21.092² + (0 × cos(37°))²) ≈ 26.5

The resultant velocity at which the water will reach the ground, v ≈ 26.5 m/s.

5 0

2 years ago

When atoms of an element are excited, they emit specific wavelengths of light. How is this similar to a fingerprint when Fraunho

Anika [276]

Answer:

As you may know, each element has a "fixed" number of protons and electrons.

These electrons live in elliptical orbits around the nucleus, called valence levels or energy levels.

We know that as further away are the orbits from the nucleus, the more energy has the electrons in it. (And those energies are fixed)

Now, when an electron jumps from a level to another, there is also a jump in energy, and that jump depends only on the levels, then the jump in energy is fixed.

Particularly, when an electron jumps from a more energetic level to a less energetic one, that change in energy must be compensated in some way, and that way is by radiating a photon whose energy is exactly the same as the energy of the jump.

And the energy of a photon is related to the wavelength of the photon, then we can conclude that for a given element, the possible jumps of energy levels are known, meaning that the possible "jumps in energy" are known, which means that the wavelengths of the radiated photons also are known. Then by looking at the colors of the bands (whose depend on the wavelength of the radiated photons) we can know almost exactly what elements are radiating them.

7 0

3 years ago