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Step2247 [10]
3 years ago
15

While watching a small fireworks display from a building 204 feet tall, you observe a rocket that travels from ground to eye lev

el in 6 seconds, and explodes. What is the maximum speed that you can guarantee that the rocket reached?
Physics
1 answer:
faltersainse [42]3 years ago
4 0
204 ft = 62.2 m
We know that the initial velocity of the rocket was 0, we can find its acceleration using:
s = ut + 1/2 at²
62.2 = 0.5a(6)²
a = 3.46 m/s²
Now, the final velocity of the rocket:
v = u + at
v = 3.46 x 6
v = 20.76 m/s = 68.1 ft/s
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The solution to the questions are given as

  • t=40.39 \mathrm{sec}
  • \varepsilon &=(0.12v)e^{0.057t}
  • the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}

\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$

\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}

\varepsilon &=(0.12v)e^{0.057t}

(b) Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$

\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}

c)

In conclusion, the direction of the induced current will be Counterclockwise.

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