The three main parts of an atom are protons, neutrons<span>, and </span>electrons<span>. </span>Protons<span> - have a positive charge, located in the </span>nucleus<span>, </span>Protons<span> and </span>neutrons<span> have nearly the same mass while </span>electrons<span> are much less massive. </span>Neutrons<span>- Have a negative charge, located in the </span><span>nucleus</span>
Answer:
x = 45 cm
Explanation:
Given that,
The length of a rod, L = 50 cm
Mass, m₁ = 0.2 kg
It is at 40cm from the left end of the rod.
We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.
The centre of mass of the rod is at 25 cm.
Taking moments of both masses such that,

The distance from the left end is 40+5 = 45 cm.
Hence, at a distance of 45 cm from the left end it will balance the rod.
Answer:
181.54 K
Explanation:
From gas laws, we know that v1/t1= v2/t2 where v and t represent volume and temperatures, 1 and 2 for the first and second container. Making t2 the subject of the formula then
T2=v2t1/ v1
Given information
V1 435 ml
V2 265 ml
T1 298K
Substituting the given values then
T2=265*298/435=181.54 K
To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.

Here,
m = mass
g = Acceleration due to gravity
Rearranging to find the charge,

Replacing,


Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is

Here,
n = Number of electrons
e = Charge of each electron

Replacing,


Therefore the number of electrons that reside on the drop is 