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3 years ago

When a piece of wood burns, it makes heat, light, and ash. Therefore, it is an example of what type of change

Physics

2 answers:

Nastasia [14]3 years ago

8 0

It is a chemical change

Ivahew [28]3 years ago

3 0

This is a chemical change because you cannot change the ash back to wood.

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A wire 1.0 m long experiences a magnetic force of 0.50 N due to a

NNADVOKAT [17]

Answer:

11.50

thenks me later

hehehehe how old are you

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3 years ago

Your outlets at home are rated at 120 V, i.e. the two prongs have on average a potential difference of 120V. If you transfer 2.7

hodyreva [135]

Answer:

E =230.4 MJ

Explanation:

As 1 mole of electron = 6X 10^23 particles.

charge of an electron is 1.6 X 10 ^-19 C

Finding Charge:

(6X10^23 ) (2.7)(1.6X10^-19 C)

i.e. 192 K C

now to find the energy released from electrons

V=E/q

E=V X q

i.e E = 120 V X 192 K C

E =230.4 MJ

4 0

3 years ago

WHAT DOES THE INCREASE IN TEMPERATURE INDICATE

KATRIN_1 [288]

Increase in temperature means:
- The substance is getting hotter
- It's internal energy is rising

8 0

3 years ago

Read 2 more answers

P.E

Vladimir79 [104]

Answer:

pagtalon?

Explanation:

because that had to be the answer

5 0

3 years ago

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Simora [160]

Answer:

The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

$F = k \frac{q_1 q_2}{d^2}$

where k is Coulomb's constant, $q_1$ and $q_2$ are the charges and d is the distance between the charges.

Working a little the equation, we can take:

$d^2 = k \frac{q_1 q_2}{F}$

$d = \sqrt{ k \frac{q_1 q_2}{F}}$

And this equation will give us the distance between the charges. Taking the values of the problem

$k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00$

(the force has a minus sign, as its attractive)

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 28,462,500 \ m^2}}$

$d = 5,335.026 m$

And this is the distance between the charges.

3 0

4 years ago