How do you calculate time from launch to explosion in a firework

Physics

2 answers:

xenn [34]3 years ago

8 0

Answer:

Explanation:

If a rocket (firework) is launched

with an initial velocity of 39.2 meters per second at a height of 1.6 meters above

the ground, the equation h = -4.9t2

+ 39.2t + 1.6 represents the rocket’s height h in

meters after t seconds. The rocket will explode at approximately the highest point.

Fire official require the fireworks to explode at a height greater than 76 feet in

order for debris to land in Lake Michigan and not on the festival sight. The event

planners assured the fire officials that the fireworks would reach a height greater

than 76 feet

Sholpan [36]3 years ago

4 0

Answer:

If a rocket (firework) is launched with an initial velocity of 39.2 meters per second at a height of 1.6 meters above the ground, the equation h = -4.9t2 + 39.2t + 1.6 represents the rocket's height h in meters after t seconds. The rocket will explode at approximately the highest point.

Explanation:

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A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$