No because an atom consists of <u>two</u> main parts <em>and</em> <u>three</u> subatomic particles - protons, neutrons, electrons. Each one is smaller than an atom, therefore they are subatomic particles. An atom only requires protons and electrons to be an atom - e.g. Hydrogen has 1 proton and 1 electron. Neutrons do not affect the overall charge of the atom, and only increase the atomic mass.
Answer
given,
net charge = +2.00 μC
we know,
1 coulomb charge = 6.28 x 10¹⁸electrons
1 micro coulomb charge = 6.28 x 10¹⁸ x 10⁻⁶ electron
= 6.28 x 10¹² electrons
2.00 μC = 2 x 6.28 x 10¹² electrons
= 1.256 x 10¹³ electrons
since net charge is positive.
The number of protons should be 1.256 x 10¹³ more than electrons.
hence, +2.00 μC have 1.256 x 10¹³ more protons than electrons.
Thank you for posting your question here at brainly. Below is the answer:
sum of Mc = 0 = -Ay(4.2 + 3cos(59)) + (275)(2.1 + 3cos(59)) + M
<span>- Ay = (M + (275*(2.1 + 3cos(59)))/(4.2 + 3cos(59)) </span>
<span>sum of Ma = 0 = (-275)(2.1) - Cy(4.2 + 3cos(59)) + M </span>
<span>- Cy = (M - (275*2.1))/(4.2 + 3cos(59)) </span>
<span>Ay + Cy = 275 = ((M+1002.41)+(M-577.5))/(5.745) </span>
<span>= (2M + 424.91)/(5.745) </span>
<span>M = ((275*5.745) - 424.91)/2 </span>
<span>= 577.483 which rounds off to 577 </span>
<span>Is it maybe supposed to be Ay - Cy = 275</span>
<h2><u>Answer:</u></h2>
The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If the trials were run on a real track with real gliders, using stopwatches and meter sticks for measurement, the data compared by the following statements:
1. (There would be variables that would be hard to control, leading to less reliable data.)
3. (Meter sticks may lack precision or may be read incorrectly.)
4. (Real glider data may vary since real collisions may involve loss of energy.)
5. (Human error in recording or plotting the data could be a factor.)
Answer:

Explanation:
B = Magnetic field = 
d = Diameter of loop = 16 cm
r = Radius = 
i = Current
= Vacuum permeability = 
The magnetic field of a loop is given by

The current needed to produce such a field at the center of the loop is 