Unbalanced as long as it is moving up or down. Immediately after being fired the vertical force from the launch will be greater than the force of gravity. for an instant at the the beginning of the projectiles decent the force will actually be balanced and there will be no vertical movement. After that the force of gravity is greater.
tl;dr they are unbalanced
Answer:
The product of the decay its Sulfur-32
Explanation:
Phosphorus-32 ( lets write it 
, where the number above its the atomic mass and the number below the atomic number) decays turning a neutron into a proton and emitting radiation on the form of a electron. This is the beta minus decay, and, actually, an electronic antineutrino its also produced. We can write this decay for an X isotope with a Y isotope produced as:
where 
its the electron, and 
the electronic antineutrino . We can see that the atomic number increases by one (cause a proton it produced and retained into the nucleus), and the atomic mass is approximately the same (there is a small difference between the neutron and proton mass, but its very small).
So, Phosphorus-32 (atomic number 15) will turn to an element with atomic number 16, and atomic mass 32, as:
.
.
The Y isotope must have an atomic number of 16 and an atomic mass of 32. The element with atomic number 16 its Sulfur (S), so, our decay its
.
and the product of such decay its Sulfur-32
Answer:
Vf = 75.4 m/s
Explanation:
In order to find the final velocity of the penny when it would hit the ground, we will use the equation of motion. In this particular case the third equation of motion can be used. The third equation of motion is written as follows:
2gh = Vf² - Vi²
where,
Vf = Final Velocity of the penny when it would hit the ground = ?
Vi = Initial Velocity of the penny = 0 m/s (Since, the penny starts from rest)
g = acceleration due to gravity = 9.8 m/s²
h = height of building = 290 m
Therefore,
2(9.8 m/s²)(290 m) = Vf² - (0 m/s)²
Vf = √(5684 m²/s²)
<u>Vf = 75.4 m/s</u>
Answer:
A. The potential energy of the positively charge proton from 5v to 7.5v will increase because in an electric field work is done on the proton to move it to a point of higher potential
B. when an electron is moved from 7.5v to 5v it loses potential energy due to the charges lost so as to bring it to a point of lower potential
Explanation:
