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4 years ago

Write a paragraph using matter, atom, nucleus, proton, neutron and electron

2 answers:

8 0

An atom is made up of different parts. The main part of the atom is the nucleus, which contains protons and neutrons. The outside of the atom is made up of rings that contain electrons. The number of protons, neutrons and epectrons can be found next to the element in the periodic table. (sorry I was a little unsure what to say about matter)

UNO [17]4 years ago

4 0

Atoms are the basic unit of matter, it cannot be broken down into anything smaller. Matter is anything that has mass and takes up space. Protons, neutrons and electrons all are particles. The nucleus is where the protons and neutrons are located, while the electrons are located in the orbitals.

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The sound intensity at a distance 2.00 m from a sound source is 5.00 Find the total sound energy emitted by the source in each s

notka56 [123]

Answer:

P = 251, 3 W

Explanation:

The intensity is defined as the power emitted per unit area

I = P / A

Since sound is distributed in all directions spherical shape, the area of a sphere is

A = 4π r²

let's clear the power and replace

P = I A

P = I (4π r²)

let's calculate

P = 5.00 (4π 2²)

P = 251, 3 W

6 0

4 years ago

An object with a mass of 20 kg has a net force of 80 N acting on it. What is the acceleration of the object?

Archy [21]

Answer:

4m/s^2

Explanation:

mass(m)=20 kg

force=80 N

acceleration (a)=?

Therefore,

Force = mass * acceleration

80 = 20*a

a=80/20

=4m/s^2

7 0

2 years ago

20 characters or more

andreev551 [17]

Answer:

yes 20 characters or more

Explanation:

3 0

3 years ago

4. During which three months is the difference the same between average high temperature and average low temperature?

kkurt [141]

Answer:

1 Inch and 20 F

Explanation:

6 0

3 years ago

wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the

lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

$E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)$

The electric field due to charge QB at P is EB leftwards

$E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)$

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0

3 years ago