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Kryger [21]
3 years ago
8

What is the initial velocity of a go-kart traveling at a uniform acceleration of 0.5 m/s^2 for 5s as it slows down to a stop?

Physics
1 answer:
Arlecino [84]3 years ago
7 0

The go-kart's velocity v after time t is given by

v=v_0+at

where v_0 is its initial velocity and a is its acceleration. After t=5\,\mathrm s, the go-kart stops completely, so

0\,\dfrac{\mathrm m}{\mathrm s}=v_0+\left(-0.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(5\,\mathrm s)\implies v_0=2.5\,\dfrac{\mathrm m}{\mathrm s}

where a because we know the go-kart is slowing down.

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Answer:

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What does the term “energy crisis” mean?
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3 years ago
an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le
storchak [24]
<span>On what:

f (is the focal length of the lens) = ? 
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}
\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}
\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}
\frac{1}{f} =  \frac{10}{33.18}
Product of extremes equals product of means:
10*f = 1*33.18
10f = 33.18
f =  \frac{33.18}{10}
\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark

6 0
3 years ago
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