Gravitational potential energy can be described as m*g*h (mass times gravity times height).
Originally,
15kg * 9.8m/s^2 *0.3 m = 44.1 kg*m^2/s^2 = 44.1 Joules.
After it is moved to a 1m shelf:
15kg * 9.8m/s * 1 = 147 kg*m^2/s^2= 147 Joules.
To find how much energy was added, we subtract final energy from initial energy:
147 J - 44.1 J = 102.9 Joules.
A) The answer is 11.53 m/s
The final kinetic energy (KEf) is the sum of initial kinetic energy (KEi) and initial potential energy (PEi).
KEf = KEi + PEi
Kinetic energy depends on mass (m) and velocity (v)
KEf = 1/2 m * vf²
KEi = 1/2 m * vi²
Potential energy depends on mass (m), acceleration (a), and height (h):
PEi = m * a * h
So:
KEf = KEi + <span>PEi
</span>1/2 m * vf² = 1/2 m * vi² + m * a * h
..
Divide all sides by m:
1/2 vf² = 1/2 vi² + a * h
We know:
vi = 9.87 m/s
a = 9.8 m/s²
h = 1.81 m
1/2 vf² = 1/2 * 9.87² + 9.8 * 1.81
1/2 vf² = 48.71 + 17.74
1/2 vf² = 66.45
vf² = 66.45 * 2
vf² = 132.9
vf = √132.9
vf = 11.53 m/s
b) The answer is 6.78 m
The kinetic energy at the bottom (KE) is equal to the potential energy at the highest point (PE)
KE = PE
Kinetic energy depends on mass (m) and velocity (v)
KE = 1/2 m * v²
Potential energy depends on mass (m), acceleration (a), and height (h):
PE = m * a * h
KE = PE
1/2 m * v² = m * a * h
Divide both sides by m:
1/2 * v² = a * h
v = 11.53 m/s
a = 9.8 m/s²
h = ?
1/2 * 11.53² = 9.8 * h
1/2 * 132.94 = 9.8 * h
66.47 = 9.8 * h
h = 66.47 / 9.8
h = 6.78 m
Answer:
C)T
Explanation:
The period of a mass-spring system is:
As can be seen, the period of this simple harmonic motion, does not depend at all on the gravitational acceleration (g), neither the mass nor the spring constant depends on this value.
These are the most common type of faults not just inductors but also with other elements too like resistors,transformers, generators etc.
open circuit fault means the flow of current is disrupted some how in the circuit and the circuit stops operating. and for short circuit fault the current in the system will be pretty high and this short circuit current or fault current will always run back to the fault location, if the inductor got short circuited somehow then the fault current will only run through it because it will then provide a very low impedence path
