Kinetic energy and potential energy pair is the quantity in which one will increase then other will decrease
As we know that sum of kinetic energy and potential energy will always remain conserved
So here we will have
so here as we move away from mean position the kinetic energy will decrease while at the same time potential energy will increase.
So the pair of potential energy and kinetic energy will satisfy the above condition
Answer:
v = 10 m/s
Explanation:
Let's assume the wheel does not slip as it accelerates.
Energy theory is more straightforward than kinematics in my opinion.
Work done on the wheel
W = Fd = 45(12) = 540 J
Some is converted to potential energy
PE = mgh = 4(9.8)12sin30 = 235.2 J
As there is no friction mentioned, the remainder is kinetic energy
KE = 540 - 235.2 = 304.8 J
KE = ½mv² + ½Iω²
ω = v/R
KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²
v = √(2KE / (m + I/R²))
v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6
v = 10.07968...
Answer:
iEvaluate for \(x=2.\)Evaluate for \(x=2.\)Evaluate for \(x=2.\)Evaluate for \(x=2.\)Evaluate for \(x=2.\)Evaluate for \(x=2.\)Evaluate for \(x=2.\)
Explanation:
No, it couldn't be.
On that scale, Neptune would be almost 1,740 MILES from the sun.
ON THAT SCALE !
