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3 years ago

A bottle rocket takes off with a = 34.5 m/s2. It

Physics

1 answer:

sergiy2304 [10]3 years ago

8 0

Answer:3.4 seconds

Explanation:

Initial velocity(u)=0

acceleration=34.5m/s^2

Height(h)=200m

Time =t

h=u x t - (gxt^2)/2

200=0xt+(34.5xt^2)/2

200=34.5t^2/2

Cross multiply

200x2=34.5t^2

400=34.5t^2

Divide both sides by 34.5

400/34.5=34.5t^2/34.5

11.59=t^2

t^2=11.59

Take them square root of both sides

t=√(11.59)

t=3.4 seconds

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Read 2 more answers

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It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

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In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be: