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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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SVETLANKA909090 [29]

Answer: (Sorry, but I don't know how to calculate mass)

1. 15 N

2. 0.4921 $\frac{ft}{s^2}$ (feet per second squared)

4. 150 N

5. 8.202 feet per second squared

3 0

2 years ago

Light travels at 3 × 108 m/s, and it takes about 8 min for light from the sun to travel to Earth. Based on this, the order of ma

N76 [4]

Answer:

The order of magnitude of the distance from the sun to Earth is 10⁸ km.

Explanation:

The order of magnitude of the distance from the sun to Earth can be calculated as follows:

$c = \frac{x}{t}$

Where:

c: is the speed of light = 3x10⁸ m/s

t: is the time = 8 min

Hence, the distance is:

$x = c*t = 3 \cdot 10^{8} m/s*8 min*\frac{60 s}{1 min} = 1.44 \cdot 10^{11} m = 1.44 \cdot 10^{8} km$

Therefore, the order of magnitude of the distance from the sun to Earth is 10⁸ km.

I hope it helps you!

5 0

3 years ago

Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at

sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

E = k*Q / r^2

Where, k: Coulomb's Constant

Q: The charge of particle

r: The distance from source

- The Electric Field due to charge 1:

E_1 = k*Q_1 / r^2

E_1 = k*Q / (2*a)^2

E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

E_2 = k*Q_2 / r^2

E_2 = k*Q_2 / (13*a)^2

E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

E_net = E_1 + E_2

2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

1: 2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

2Q = Q / 4 + Q_2 / 169

Q_2 = 295.75*Q

2: 2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

2Q = Q / 4 - Q_2 / 169

Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0

3 years ago

What crop is least likely to do well when the temperatures are very hot?

torisob [31]

a. Sweet corn and possibly d. okra.

3 0

3 years ago

How far can light travel in one year?

elena-14-01-66 [18.8K]

The answer is b
300,000 km

8 0

3 years ago