A truck is traveling 20m/s accelerates 3 m/s^2 for 4 seconds how far did it travel while it was accelerating. Using guess method

Physics

1 answer:

stellarik [79]3 years ago

5 0

<h3> $\huge\underline\bold\blue{ƛƝƧƜЄƦ}$ </h3><h3>Given</h3>

$\blue\star$ v = 20m\s

$\blue\star$ a = 3m\s^2

$\blue\star$ t = 4sec

Firstly we have to find u

$\star$ a = $\dfrac{v - u}{t}$

$\star$ 3m\s = $\dfrac{20 - u}{4}$

$\star$ 12m\s = 20 - u

$\star$ 20 - u = 12m\s

$\star$ - u = -8

$\star$ u = 8

Now we can easily find distance by using second equation of motion

$\red\star$ s = ut + 1\2 at^2

$\red\star$ s = 8(4) + 1\2(3)(16)

$\red\star$ s = 32 + 24

$\red\star$ s = 56

So distance is 56 m\s hope it helps

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Suppose that a star has a spectrum that includes red, blue, and violet lines spaced in the pattern of the lines from hydrogen bu

ladessa [460]

Answer:

It can be concluded that the star is moving away from the observer.

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 is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest for this case is 434 nm and 410 nm ( $\lambda_{0} = 434nm$ , $\lambda_{0} = 410nm$ )

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

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

Since, $\lambda_{measured}$ (444nm) is greater than $\lambda_{0}$ (434 nm) and $\lambda_{measured}$ (420nm) is greater than $\lambda_{0}$ (410 nm), it can be concluded that the star is moving away from the observer