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2 years ago

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Astronomers arrange the stars into groups called spectral classes (or types) according to the kinds of lines they find in their

spectra. These spectral classes are arranged in order of:

1 answer:

MArishka [77]2 years ago

8 0

Answer:

Temperature

Explanation:

The spectral types are:

O, B, A, F, G, K, M

They are arranged in order from the highest temperature to the coolest temperature

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A particle moving along the x-axis has a position given by x = (24t – 2.0t 3 ) m, where t is measured in s. What is the magnitud

Vitek1552 [10]

<h2>The magnitude 24 ( $\dfrac{m}{s^2}$ ) of the acceleration of the particle when the particle is not moving.</h2>

Explanation:

Given,

A particle moving along the x-axis has a position given by

$x=(24t-2.0t^3)$ m ........ (1)

To find, the magnitude ( $\dfrac{m}{s^2}$ ) of the acceleration of the particle when the particle is not moving = ?

Differentiating equation (1) w.r.t, 't', we get

$\dfrac{dx}{dt} =\dfrac{d((24t-2.0t^3))}{dt}$

⇒ $\dfrac{dx}{dt} =24(1)-3(2.0)t^{2} =24-6t^{2}$ ....... (2)

⇒ $24-6t^{2} = 0$

⇒ $t^{2}=2^{2}$

⇒ t = 2 s

Again, differentiating equation (2) w.r.t, 't', we get

$\dfrac{d^2x}{dt^2} =-12t$

Put t = 2, we get

$\dfrac{d^2x}{dt^2} =-12(2)=24$

Thus, the magnitude 24 ( $\dfrac{m}{s^2}$ ) of the acceleration of the particle when the particle is not moving.

3 0

3 years ago

Sedimentary rocks are the only rocks that can potentially containa.fossils.c.fractures.b.minerals.d.faults. Please select the be

Elena-2011 [213]

Hello <span>Mr1guy24 </span>

Question: <span>Sedimentary rocks are the only rocks that can potentially contain?
</span>
Answer: Fossils (A)

Reason: This is what makes sedimentary rocks unique

Hope This Helps!
-Chris

8 0

3 years ago

Read 2 more answers

Which of the statements concerning light are true? The speed of light is the same no matter what material it is traveling throug

wel

Answer:

The statements that are true concerning light are the last three statements:

Its propagation direction is perpendicular to both the electric field and the magnetic field.

It moves at a constant speed through a vacuum.

The speed of light in matter is less than the speed of light in a vacuum.

Explanation:

<em>Light</em> is <em>electromagnetic waves. </em>

The properties of the electromagnetic waves were established by James Clerk Maxwell.

They included that they are the result of the oscillation of a <em>magnetic field </em>in phase with an <em>electric field</em> which are always is always <em>perpendicular</em> to each other.

Also, the electromagnetic waves propagate at right-angles to the direction of both the magnetic and the electric field, meaning that they are a type of transverse wave.

Thus, the second statement (<em>"Its propagation direction is parallel to both the electric field and the magnetic field"</em>) is false, and the fourth statement ("Its propagation direction is perpendicular to both the electric field and the magnetic field") is true.

On the other hand, it is a postulate of the special theory of relativity that the speed of light is a constant (absolute value) in vacuum: nothing can travel faster than what light travels in vacuum. Thus, the fifth statement, <em>"It moves at a constant speed through a vacuum"</em> is true.

About the speed of light in matter, it is always less than the speed of light in vacuum. Thus, the first statement, "<em>the speed of light is the same no matter what material it is traveling through</em>", and the third statement "<em>the speed of light in matter is greater than the speed of light in a vacuum"</em> are false; while the last statement, "<em>the speed of light in matter is less than the speed of light in a vacuum</em>" is true.

The explanation on why the speed of light is less in a medium than in vacuum is related with the fact that at nanoscopic level the waves suffer polarization which means deviations from the straighi path, which makes that the net straight propagation is slower.

8 0

3 years ago

1. Al descomponerse su vehículo una persona tira de su auto con la ayuda de una cuerda con una fuerza de 3500 N que forma un áng

OleMash [197]

Responder:

Fy = 2474,8737

Fx = 2474,8737

Explicación:

Dado que :

Dado:

Fuerza, F = 3500 N

Ángulo formado con la horizontal, θ, = 45 °

Los componentes de una fuerza se pueden descomponer en componentes verticales y horizontales.

El componente vertical Fy; y

El componente horizontal Fx

Fy = Fuerza * sinθ

Fy = 3500 * sin45 °

Fy = 2474,8737

El componente horizontal:

Fx = Fuerza * cosθ

Fy = 3500 * cos45 °

Fy = 2474,8737

8 0

2 years ago

Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo

Elena-2011 [213]

a) $3.14 \cdot 10^{-4} s$

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

a)

The position of the tip of the lever at time t is described by the equation:

$y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]$ (1)

The generic equation that describes a wave is

$y(t)=A sin (\frac{2\pi}{T} t)$ (2)

where

A is the amplitude of the wave

T is the period of the wave

t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

$\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}$

Therefore, the period is

$T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s$

b)

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when $sin(...)=1$ , and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

c)

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

$\lambda=2L$

where

$\lambda$ is the wavelength of the wave

L is the length of the string

In this problem,

L = 5.00 m is the length of the string

Therefore, the wavelength is

$\lambda =2(5.00)=10.0 m$

d)

The amplitude of a wave is the magnitude of the maximum displacement of the wave, measured relative to the equilibrium position.

In this problem, we can easily infer the amplitude of this wave by looking at eq.(1).

$y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]$

And by comparing it with the general equation of a wave:

$y(t)=A sin (\frac{2\pi}{T} t)$

In fact, the maximum displacement occurs when the sine part is equal to 1, so when

$sin(\frac{2\pi}{T}t)=1$

which means that

$y(t)=A$

And therefore in this case,

$y=0.500 cm$

So, this is the displacement.

6 0

2 years ago