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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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A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will i
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Answer:

350 F to 100 F it take approx 87.33 min  

Explanation:

given data

oven = 350◦F

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solution

we apply here Newtons law of cooling  

\frac{dT}{dt} = -k(T-Ta)

\frac{dy}{dt} = \frac{d}{dt} (T(t) -Ta)

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put her value for time 30 min and T(t) = 200◦F and To =350◦F  and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) e^{-k30}

k = 0.025575

so here for  T(t) = 100F

100 = 70 + ( 350 - 70 ) e^{-0.025575*t}

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min  

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