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3 years ago

What is the average speed of a car that moves 570 m in 24 s

1 answer:

7 0

The formula for speed is:
Speed = Distance/Time
We can plug in the given values into the above equation:
Speed = 570m÷24s
Speed = 23.75, which rounds to 24m/s as a whole number. Therefore, the answer is b.

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A charge moves from pount A to point B in an electric field. what is true about the potential energy of the charge?

iren [92.7K]

I'm going to have to say its B

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3 years ago

Read 2 more answers

Dirt can increase friction and cause the worker to slip a)-True b)- false

iVinArrow [24]

Answer:A}

Explanation:Because it happened to me lol

6 0

3 years ago

A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N

Sergio [31]

Answer:

200cm

Explanation:

Answer:

100cm

Explanation:

Using

F= ( N/2L)(√T/u)

F1 will now be (0.5*2)( √600/0.015)

=> L( wavelength)= 200/2cm = 100cm

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3 years ago

Imagine that you have obtained spectra for several galaxies and have measured the observed wavelength of a hydrogen emission lin

Anna71 [15]

Answer:

Galaxy 1:

z = 0.0056

Galaxy 2:

z = 0.014

Galaxy 3:

z = 0.040

Explanation:

Spectral lines will be shifted to the blue part of the spectrum¹ if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect). The source in this particular case is represented for each of the galaxies of interest.

Hence, the redshift represents this shift of the spectral lines to red part in the spectrum of a galaxy or any object which is moving away. That is a direct confirmation of how the universe is in an expanding accelerated motion.

The redshift can be defined in analytic way by means of the 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.

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

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

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

Where z is the redshift.

For the case of Galaxy 1:

Where $\lambda_{measured} = 660.0 nm$ and $\lambda_{0} = 656.3 nm$ .

$z = \frac{\lambda_{measured}-\lambda_{0}}{\lambda_{0}}$

$z = (\frac{660.0 nm - 656.3 nm}{656.3 nm})$

$z = 0.0056$

For the case of Galaxy 2: 

Where $\lambda_{measured} = 665.8 nm$ and $\lambda_{0} = 656.3 nm$ .

$z = \frac{665.8 nm - 656.3 nm}{656.3 nm}$

$z = 0.014$

For the case of Galaxy 3:

Where $\lambda_{measured} = 682.7 nm$ and $\lambda_{0} = 656.3 nm$ .

$z = \frac{682.7 nm - 656.3 nm}{656.3 nm}$

$z = 0.040$

Key terms:

¹Spectrum: Decomposition of light in its characteristic colors (wavelengths).

8 0

3 years ago

How can red giants be so bright when they are so cool

Norma-Jean [14]

Answer:

The star's luminosity rises above its previous level. Because it is so cool, the surface will be red, and it will be much farther away from the center than it was during the earlier stages of star evolution. Despite its lower surface temperature, the red giant has a large surface area, which makes it very luminous.

7 0

2 years ago