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3 years ago

The spectra of most stars are dark-line spectra because ________.

1 answer:

5 0

Because dark line spectra result from passing white light through ionized gasses and plasmas, which is what the atmosphere of stars are made of. These frequencies are scattered by the star's atmosphere as it leaves the surface (photosphere) of the star, and don't make it to earth.

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10) Two students want to use a 12-meter long rope to create standing waves. They first measure the speed at which a single wave

zhannawk [14.2K]

Answer:

To create a second harmonic the rope must vibrate at the frequency of 3 Hz

Explanation:

First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,

f₁ = v/2L

where,

v = speed of wave = 36 m/s

L = Length of rope = 12 m

f₁ = fundamental frequency

Therefore,

f₁ = (36 m/s)/2(12 m)

f₁ = 1.5 Hz

Now the frequency of nth harmonic is given in general, as:

fn = nf₁

where,

fn = frequency of nth harmonic

n = No. of Harmonic = 2

f₁ = fundamental frequency = 1.5 Hz

Therefore,

f₂ = (2)(1.5 Hz)

5 0

3 years ago

Because of your success in physics class you are selected for an internship at a prestigious bicycle company in its research and

tiny-mole [99]

To develop the problem it is necessary to apply the equations related to the moment of inertia.

The given values can be defined as,

$M = 1.0 kg$

$r = 0.5 m$

$m = 10 g$

$I = 0.280 kg.m^2$

According to the definition of the moment of inertia applied to the exercise we can arrive at the equation that,

$I = I_{rim} + n * I_{spoke}$

Where n is the number of spokes necessary to construct the wheel.

$I_{rim} = M*r^2 = 1.0 * 0.5^2$

$I_{spoke} = \frac{1}{3} * m * r^2 = \frac{1}{3}* 10 * 10^-3 * 0.5^2$

Replacing the values at the general equation we have,

$0.280 = 1.0 * 0.5^2 + n * (1/3 * 10 * 10^-3 * 0.5^2 )$

Solving for n,

$n = 36$

Therefore the number of spokes necessary to construct the wheel is 36

PART B) The mass of the wheel is given by the sum of all masses and the total spokes, then

$M_w= M + n*m$

$M_w = 1.0 + 36* 10 * 10^{-3} Kg$

$M_w = 1.36 Kg$

Therefore the mass of the wheel must be of 1.36Kg

4 0

4 years ago

Which statement correctly stated kepler's 3rd law of planetary motion?

nadya68 [22]

When you square the "year" of each planet and divide it by the cube of its distance, or axis from the sun, the number would be the same for all the planets

8 0

4 years ago

Read 2 more answers

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$

where

m = 950 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\Delta h$ is the change in height, which is

$\Delta h = L sin 30^{\circ}$

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

$\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J$

8 0

3 years ago

A mountaintop is a height y above the level ground. A woman measures the angle of elevation of the mountaintop to be θ when she

valentina_108 [34]

Let t = Theta and p = Phi
Tan t = y/x Then x =y/Tant.
Tant = y/(x-d) x-d = y/Tanp
y/Tant - d = y/Tanp
y -d*Tanr = y*Tant/Tanp
y-y*Tant/Tanp = d*Tanr
y(1 - Tanr/Tanp = d*Tant

y = d*Tant/(1-Tant/Tanp)

7 0

4 years ago