Answer:
B. By adding the number of protons and the number of neutrons
Explanation:
The atomic mass is determined by adding the number of protons and neutrons in an atom. An atom is made up of three fundamental particles: Electrons, Protons and Neutrons.
The protons and neutrons occupy a central region in an atom known as the nucleus. The nucleus is positively charged and mass concentrated.
If we compare the relative masses of the subatomic particles, the masses of protons and neutrons would be 1 and that of an electron would be 1/1840. This shows that the mass of electrons are negligible.
In order to ascertain atomic mass, we therefore add the number of protons and neutrons together. This is how we arrive at 12.011 as the value of the atomic mass of C and for other elements.
The atomic mass is also known as the mass number.
<span>Acceleration is the rate of
change of the velocity of an object that is moving. This value is a result of
all the forces that is acting on an object which is described by Newton's
second law of motion. Calculations of such is straightforward, if we are given
the final velocity, the initial velocity and the total time interval. However, we are not given these values. We are only left by using the kinematic equation expressed as:
d = v0t + at^2/2
We cancel the term with v0 since it is initially at rest,
d = at^2/2
44 = a(6.2)^2/2
a = 2.3 m/s^2
</span>
Given data:
* The mass of the ball is 2 kg.
* The gravitational field strength at the surface of planet X is 5 N/kg.
Solution:
The weight of the ball on the planet X is,
where m is the mass of ball, a is the gravitational field strength,
Substituting the known values,
Thus, the weight of the ball on the surface of planet X is 10 N.
The Swamp
There are many websites that say the rainforest but the rainforest is warmer and gets rain year round and swamps are warm and gets lots of rain but not year round.
Answer:
84.82N/C.
Explanation:
The x-components of the electric field cancel; therefore, we only care about the y-components.
The y-component of the differential electric field at the center is
.
Now, let us call 
the charge per unit length, then we know that
;
therefore,
Integrating
![$E = \frac{k \lambda }{R}*[-cos(\pi )+cos(0) ]$](https://tex.z-dn.net/?f=%24E%20%3D%20%5Cfrac%7Bk%20%5Clambda%20%20%20%7D%7BR%7D%2A%5B-cos%28%5Cpi%20%29%2Bcos%280%29%20%5D%24)
Now, we know that
and the radius of the semicircle is
therefore,
