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Alona [7]
3 years ago
5

Redshift means stars and galaxies are emitting a color that is shifting toward the red end of the color spectrum. this indicates

the light's wavelengths are _____ and less frequent, and the star or galaxy is____?
Physics
2 answers:
iren [92.7K]3 years ago
7 0
Increasing ; moving away

This indicates the light's wavelengths are (increasing) and the star or galaxy is (moving away.)
Olin [163]3 years ago
6 0

Answer:

Redshift means stars and galaxies are emitting a color that is shifting toward the red end of the color spectrum. this indicates the light's wavelengths are <u>MORE</u> and less frequent, and the star or galaxy is <u>MOVING AWAY</u>

Explanation:

As we know that by Doppeler's effect of light when source is moving away from the observer then the frequency observed by the observer is given as

\frac{\Delta \nu}{\nu} = \frac{v}{c}

so here on moving away the frequency observed by observer will decrease and hence we can say that the wavelength of light must be increased

So here on increasing the observed wavelength we say it to be shifted towards higher side which means it is red shift

so here correct answer would be

Redshift means stars and galaxies are emitting a color that is shifting toward the red end of the color spectrum. this indicates the light's wavelengths are <u>MORE</u> and less frequent, and the star or galaxy is <u>MOVING AWAY</u>

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How are science and technology the same
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Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a
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Answer:

The maximum speed at which the car can safety travel around the track is 18.6m/s.

Explanation:

Since the car is in circular motion, there has to be a centripetal force F_c. In this case, the only force that applies for that is the static frictional force f_sbetween the tires and the track. Then, we can write that:

f_s=F_c

And since f_s\leq \mu N and F_c=\frac{mv^{2}}{r}, we have:

\mu N\geq \frac{mv^{2}}{r}

Now, if we write the vertical equation of motion of the car (in which there are only the weight and the normal force), we obtain:

N-mg=0\\\\\\implies N=mg

Substituting this expression for N and solving for v, we get:

\mu mg\geq \frac{mv^{2}}{r}\\\\v\leq \sqrt{\mu gr}

Finally, plugging in the given values for the coefficient of friction and the radius of the track, we have:

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Answer:

a) 4.0 rad/s2

Explanation:

  • For rigid bodies, Newton's 2nd law becomes :

       τ = I  * α (1)

        where τ is the net external torque applied, I is the rotational inertia

       of  the body with respect to the axis of rotation, and α is the angular

       acceleration caused by the torque.

  • At the same time, we can apply the definition of torque to the left side of (1), as follows:

       \tau = F*r*sin \theta (2)

       where τ = external net torque applied by Fnet, r is the distance    

       between the axis of rotation and the line of Fnet, and θ is the

      angle between both vectors.

      In this particular case, as Fnet  is applied tangentially to the disk, Fnet

     and r are perpendicular each other.

  • Since left sides of (1) and (2) are equal each other, right sides are equal too, so we can solve for the angular acceleration as follows:

       \alpha = \frac{F*r}{I} = \frac{5.0N*1.6m}{2.0 kg*m2} = 4.0 rad/s2  (3)        

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