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Scorpion4ik [409]
3 years ago
9

A EXAMPLE OF WHEN BOTH PHYSICAL CHANGE AND CHEMICAL CHANGE OCCUR 3 EXAMPLES

Physics
1 answer:
kvasek [131]3 years ago
5 0

Answer:

I know 1, that is in the case of a burning of a candle.

Explanation:

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What is this object? Explain how it works and which one of Newton’s laws this represents.
ryzh [129]

Answer:

You never gave a paper I want to see the sheet pls then maybe I will see and help you did not explain it really well

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3 years ago
Which particle is a negatively charged ion?
wel

Answer:

option D is correct

Explanation:

cl has 17 electrons if it has 18 it becomes cl-

6 0
2 years ago
Pls help on this one?
sattari [20]
The answer is point C
4 0
3 years ago
A 4.33 kg cat has 41.7 J of KE How fast is the cat moving?
balandron [24]

Answer:

The answer to your question is:

Explanation:

Data

mass = 4.33 kg

E = 41.7 J

v = ?

Formula

Ke = (1/2)mv²

Clear v from the equation

v = √2ke/m

Substitution

v = √2(41.7)/4.33

v = 19.26 m/s          Result

7 0
3 years ago
In the steady state 1.2 ✕ 1018 electrons per second enter bulb 1. There are 6.3 ✕ 1028 mobile electrons per cubic meter in tungs
bekas [8.4K]

Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

V=\frac{I}{nAq}

Where,

V= Drift Velocity

I= Flow of current

n= number of electrons

q = charge of electron

A = cross-section area.

For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is

\frac{I}{q} = 1.2*10^{18}

A= 1.3*10^{-8}m^2

n=6.3*10^{28} e/m^3

\omicron{O} = 1.2*10^{-4}(m/s)(N/c) Mobility

We can find the drift velocity replacing,

V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}

V= 1.465*10^-3m/s

The electric field is given by,

E= \frac{V}{\omicron{O}}

E=\frac{1.465*10^-3}{1.2*10^{-4}}

E=12.2V/m

7 0
3 years ago
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