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3 years ago

What type of energy is released when a firecracker explodes

Physics

2 answers:

Afina-wow [57]3 years ago

6 0

There is heat, light and sound energy, that is released.

Ludmilka [50]3 years ago

5 0

-- Sound energy. That's how you hear it explode.

-- Light energy. That's how you see it in the sky.

-- Heat energy. That's why you get burned if sparks fall on you.

-- Potential energy, as all the pieces fall to the ground.

These were all stored in it as chemical energy before it exploded.

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Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou

Gennadij [26K]

Answer:

a. The station is rotating at $1.496 \frac{rev}{min}$

b. the rotation needed is $2.8502 \frac{rev}{min}$

Explanation:

We know that the centripetal acceleration is

$a_{c}= \omega ^2 r$

where $\omega$ is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

The rotational speed is :

$2.7 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.02454 \frac{rad^2}{s^2} }$

$\omega = 0.1567 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.1567 \frac{rad}{s} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 1.496 \frac{rev}{min}$

The rotational speed needed is :

$9.8 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.08909 \frac{rad^2}{s^2} }$

$\omega = 0.2985 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.2985 \frac{rev}{min} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 2.8502 \frac{rev}{min}$

3 0

3 years ago

Read 2 more answers

Why would you expect the speed of light to be slightly less in the atmosphere then in a vacuum?

azamat

The speed of light to be slightly less in atmosphere then in vacuum because of absorption and re-emission of light by the atmospheric molecules occurred when light travels through a material

<u>Explanation:</u>

When light passes through atmosphere, it interacts or transmits through the transparent molecules in atmosphere. In this process of transmission through atmosphere, the light will be getting absorbed by them and some will get re-emitted or refracted depending upon wavelength.

But in vacuum the absence of any kind of particles will lead to no interaction and no energy loss, thus the speed of the light will be same in vacuum while due to interactions with molecules of atmosphere, there speed will be slightly less compared to in vacuum.

7 0

3 years ago

What is the safety rules?

Black_prince [1.1K]

Answer:

Quick question do you mean what are some safety rules

Explanation:

Crosswalk, Stop sign,

5 0

3 years ago

Read 2 more answers

Learning Goal:

enot [183]

Answer:

A. $U_0 = \dfrac{\epsilon_0 A V^2}{2d}$

B. $U_1 = \dfrac{\epsilon_0 A V^2}{6d}$

C. $U_2 = \dfrac{K\epsilon_0 A V^2}{2d}$

Explanation:

The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

$C=\dfrac{\epsilon A}{d}$

$C$ is the capacitance, $A$ is the common plate area, $d$ is the plate separation and $\epsilon$ is the permittivity of the material between the plates.

For air or free space, $\epsilon$ is $\epsilon_0$ called the permittivity of free space. In general, $\epsilon=\epsilon_r \epsilon_0$ where $\epsilon_r$ is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, $\epsilon_r=1$ .

The energy stored in a capacitor is the average of the product of its charge and voltage.

$U = \dfrac{QV}{2}$

Its charge, $Q$ , is related to its capacitance by $Q=CV$ (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for $U$ ,