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3 years ago

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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The human ear canal is about 2.6 cm long and can be regarded as a tube open at one end and closed at the eardrum. What is the fu

Ugo [173]

Answer:

f = v/λ

= v/4*L

= 343 / 4 (0.026)

= 3120 Hz

8 0

3 years ago

Read 2 more answers

Two cars cover the same distance in a straight line. Car a covers the distance at a constant velocity. Car b starts from rest an

Artyom0805 [142]

a) For the motion of car with uniform velocity we have , $s = ut+\frac{1}{2}at^2$ , where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.

In this case s = 520 m, t = 223 seconds, a =0 $m/s^2$

Substituting

$520 = u*223\\ \\u = 2.33 m/s$

The constant velocity of car a = 2.33 m/s

b) We have $s = ut+\frac{1}{2} at^2$

s = 520 m, t = 223 seconds, u =0 m/s

Substituting

$520 = 0*223+\frac{1}{2} *a*223^2\\ \\ a = 0.0209 m/s^2$

Now we have v = u+at, where v is the final velocity

Substituting

v = 0+0.0209*223 = 4.66 m/s

So final velocity of car b = 4.66 m/s

c) Acceleration = 0.0209 $m/s^2$

7 0

3 years ago

4 facts about relative dating?

sasho [114]

Relative dating is used to arrange geological events….

Relative dating puts geologic events in chronological order without requiring that a specific numerical age be assigned to each event….

Relative Dating uses the half life of isotopes to get the exact age of a rock or mineral.

6 0

2 years ago

A 40-cmcm-long tube has a 40-cmcm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. A

Licemer1 [7]

Answer:

1070 Hz

Explanation:

First, I should point out there might be a typo in the question or the question has inconsistent values. If the tube is 40 cm long, standing waves cannot be produced at 42.5 cm and 58.5 cm lengths. I assume the length is more than the value in the question then. Under this assumption, we proceed as below:

The insert in the tube creates a closed pipe with one end open and the other closed. For a closed pipe, the difference between successive resonances is a half wavelength $\frac{\lambda}{2}$ .