B. Kinetic energy increases, potential energy decreases.
The number of protons always differs in atoms of different elements.
Answer:
However, the disadvantages are:
1. Many atimes for some motion prolems, free-body diagrams has to be drawn many times so to have enough equations to solve for the unknowns. This is not the same with energy conservation principles.
2. In situations where we need to find the internal forces acting on an object, we can't truly solve such problems using free-body diagram as it captures external forces. This is not the same with energy conservation principles.
Explanation:
Often times the ideal method to use in solving motion problem related questions are mostly debated.
Energy conservation principles applies to isolated systems are useful when object changes their positions in moving upward or downward converts its potential energy due to gravity for kinetic energy, or the other way round. When energy in a system or motion remains constant that is energy is neither created nor destroyed, it can therefore be easier to calculate other unknown paramters like in the motion problem velocity, distance bearing it in mind that energy can only change from one type to another.
On the other hand, free body diagram which is a visual representation of all the forces acting on an object including their directions has so many advantages in solving motion related problems which include finding relationship between force and motion in identifying the force acting on a body.
Energy
Waves carry energy as they travel.
The question is incomplete, the complete question is;
A spherical steel ball bearing has a diameter of 2.540cm at 25.00∘C.
(a) What is its diameter when it's temperature is raised to
100∘C?
(b) What temperature change is required to increase its volume by
1.000% ?
Answer:
a) 2.542 cm
b) 303.03°C
Explanation:
Given;
Diameter of the ball= 2.540cm
Initial temperature= 25.0°C
Final temperature= 100.0°C
Percentage increase in volume = 1.000%
Temperature coefficient of expansion for steel =11.0×10^−6/∘C
d2= d1[1 + α(T2-T1)]
d2= 2.540[1 + 11.0×10^−6(100-25)]
d2= 2.540[1 + 8.25×10^-4]
d2= 2.542 cm
From;
%V ×1/100 = V ×3α ×∆T/ V
Substituting values;
1.000 ×1/100= 3× 11.0×10^−6 × ∆T
∆T= 0.01/3× 11.0×10^−6
∆T= 303.03°C