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2 years ago

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You have a large metal sphere that starts out isolated and has no net charge. Then you bring a positively charged rod near it wi

thout touching.
Will there be a force between the sphere and rod?

Draw a qualitative diagram showing the location of the charges in the metal sphere once the rod is brought near.

1 answer:

liq [111]2 years ago

6 0

Answer:

Yes

Explanation:

There will be a force because the large metal sphere is polarized when the rod is placed close to it, this happens because the metal sphere has free electrons and they will be attracted by the positively charged rod, the sphere will remain neutral as a whole (the same number of negative and positive charge), but a portion of the sphere will become more positive and the other more negative, creating this force.

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A particle moves along the x axis so that its velocity at time t is given by v(t)=

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Answer:

a = 0.7267 , acceleration is positive therefore the speed is increasing

Explanation:

The definition of acceleration is

a = dv / dt

they give us the function of speed

v = - (t-1) sin (t² / 2)

a = - sin (t²/2) - (t-1) cos (t²/2) 2t / 2

a = - sin (t²/2) - t (t-1) cos (t²/2)

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remember that the angles are in radians

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300 000 0 squared = 2 x 9.8 distance

KINEMATICS

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A student writes the following script for a scene in a futuristic movie.

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2 years ago

In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this

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Answer:

1). $v = - 2960526m/s$

2). Toward us

3). $v = - 493421m/s$

4). Toward us

5). $v = 1480263m/s$

6). Away from us

7). $v = 3207236m/s$

8). Away from us

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest is 121.6 nm ( $\lambda_{0} = 121.6nm$ )

$Redshift: \lambda_{measured} > \lambda_{0}$

$Blueshift: \lambda_{measured} < \lambda_{0}$

Then, for this particular case it is gotten:

Star 1: $\lambda_{measured} = 120.4nm$

Star 2: $\lambda_{measured} = 121.4nm$

Star 3: $\lambda_{measured} = 122.2nm$

Star 4: $\lambda_{measured} = 122.9nm$

Star 1:

$Blueshift: 120.4nm < 121.6nm$

Toward us

Star 2:

$Blueshift: 121.4nm < 121.6nm$

Toward us

Star 3:

$Redshift: 122.2nm > 121.6nm$

Away from us

Star 4:

$Redshift: 122.9nm > 121.6nm$

Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

$v = c\frac{\Delta \lambda}{\lambda_{0}}$ (1)

Where $\Delta \lambda$ is the wavelength shift, $\lambda_{0}$ is the wavelength at rest, v is the velocity of the source and c is the speed of light.

$v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}})$ (2)

<em>Case for star 1 $\lambda_{measured} = 120.4 nm$ :</em>

<em></em>

$v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})$

$v = - 2960526m/s$

Notice that the negative velocity means that is approaching to the observer.

<em>Case for star 2 $\lambda_{measured} = 121.4 nm$ :</em>

$v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})$

$v = - 493421m/s$

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$v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})$

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<em>Case for star 4 $\lambda_{measured} = 122.9 nm$ :</em>

$v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})$

$v = 3207236m/s$

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3 years ago

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3 years ago

Read 2 more answers