Answer:
A. <em>Loses PE and gains KE
</em>
Explanation:
Statement is incomplete. Complete statement of problem is:
<em>1. An object from a certain height falls freely. Which of the following happens to PE and KE when the object is half on its way down?
</em>
A. <em>Loses PE and gains KE
</em>
B. <em>Gains PE and loses KE
</em>
C. <em>Loses both PE and KE
</em>
D. <em>Gains both PE and KE</em>
If we neglect the effects of any conservative force, the application of the Principle of Energy Conservation is reduced to a sum of gravitational potential (
) and translational kinetic energies, measured in joules. That is:
(Eq. 1)
Let assume that an object falls from a height 
with a speed of zero. By definitions of gravitational potential and translational kinetic energies the previous is expanded. If final height is the half of initial value, then:
(Eq. 1b)
In a nutshell, translational kinetic energy is increased at the expense of diminishing gravitational potential energy. The correct answer is A.
Answer:
False
Explanation:
The charge on an electron as well as proton is same. The difference between electron and proton is that the electrons have a negative charge while the protons have positive charge.
The weight of an electron is, 
while the weight of a proton is 
.
Therefore, the given statement "A proton carries positive charge, and weights the same as an electron." is false.
Hydrogen and then its helium
Answer: 11%
Explanation:
Given that
X = ab^2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3 respectively.
Percentage error of b = 2%
Percentage error of b^2 = 2 × 2 = 4
When you are calculating for percentage error that involves multiplication and division, you will always add up the percentage error values.
Percentage error of X will be;
Percentage error of a + percentage error of b^2 + percentage error of c
Substitute for all these values
4 + 4 + 3 = 11%
Therefore, percentage error of X is 11%
