Mark places new batteries in his CD player and turns it on. Identify the correct energy transformation.

Physics

2 answers:

disa [49]3 years ago

6 0

<span>Chemical → electrical → mechanical and sound is the answer ^w^</span>

borishaifa [10]3 years ago

5 0

Answer is:
<span>Chemical → electrical → mechanical and sound
</span>
Explanation:
Let's take this step by step:
1- For the batteries to work, chemical reactions must occur within the battery itself. This is considered "chemical energy"
2- For the CD player to start operating, it needs electrical energy that is supplied from the batteries
3- When the CD player gains this electrical energy, it can then function correctly and play the songs producing mechanical and sound energies.

So, basically the flow goes as follows:
Chemical reactions within the batteries lead to the formation of electrical energy which operates the CD player producing mechanical and sound energy.

Hope this helps :)

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A child in a tree house uses a rope attached to a basket to lift a 24 N dog upward through a distance of 4.9 m into the house.

jenyasd209 [6]

<h3>Answer:</h3>

117.6 Joules

<h3>Explanation:</h3>

<u>We are given;</u>

Force of the dog is 24 N
Distance upward is 4.9 m

We are required to calculate the work done

Work done is the product of force and distance
That is; Work done = Force × distance
It is measured in Joules.

In this case;

Force applied is equivalent to the weight of the dog.

Work done = 24 N × 4.9 m

= 117.6 Joules

Hence, the work done in lifting the dog is 117.6 Joules

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3 years ago

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zhuklara [117]

Answer:

false

Explanation:

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3 0

3 years ago

You have been working on a new, strong, lightweight ceramic which will be used to replace steel bearing balls. One cubic meter o

marin [14]

Answer:

r=2.6 cm

Explanation:

Hi!

Lets call steel material 1, and the new alloy material 2. You know their densities:

$\rho_1=8.08*10^3\frac{kq}{m^3}\\\rho_2=3.14*10^3\frac{kq}{m^3}$

The volume of a sphere with radius r is given by:

$V=\frac{4}{3}\pi r^3$

Then the masses of the bearings are:

$m_1=\rho_1V_1=\frac{4}{3}\pi r_1^3 \\m_1=\rho_2V_2=\frac{4}{3}\pi r_2^3$

For the masses to be the same:

$\rho_1 V_1 = \rho_2 V_2\\\frac{V_2}{V_1} =\frac{\rho_1}{\rho_2}\\(\frac{r_2}{r_1})^3 = \frac{8.08}{3.14}=2.57\\r_2=\sqrt[3]{2.57}\;r1 =1.37r_1=1.37*1.9cm=2.6cm$