Answer:
124.52 mL
Explanation:
from Boyle's Law,
PV = P'V' ................... Equation 1
Where P = Initial pressure of the gas, V = Initial volume of the gas, P' = Final pressure of the gas, V' = Final volume of the gas.
make V' the subject of the equation.
V' = PV/P'............. Equation 2
Given: P = 267 torr = (267×0.00131) = 0.34977 atm, V = 356 mL, P' = 1 atm
Substitute into equation 2
V' = (0.34977×356)/1
V' = 124.52 mL.
Hence the new volume of the balloon = 124.52 mL
<span>Last choice on the list:
Object A has a net charge of 0 because the positive and negative
charges are balanced.
Object B has a net charge of –2 because there is an imbalance of
charged particles (2 more negative electrons than positive protons).</span>
At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:

while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:

And the total energy of the system is

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:

while the mass is moving at speed v, and therefore the kinetic energy is

And the total energy is

For the law of conservation of energy, the total energy must be conserved, therefore

. So we can write

that we can solve to find an expression for v:
Answer:
54%
Explanation:
So, we have that the "magnitude of its displacement from equilibrium is greater than (0.66)A—''. Thus, the first step to take in answering this question is to write out the equation showing the displacement in simple harmonic motion which is = A cos w×t.
Therefore, we will have two instances t the displacement that is to say at a point 2π/w - a2 and the second point at a = a2.
Let us say that 2π/w = A, then, we have that a = A cos ^-1 (0.66)/2π. Also, we have that a2 = A/2 - A cos^- (0.66) / 2π.
The next thing to do is to calculate or determine the total length of of the required time. Thus, the total length is given as:
2a1 + ( A - 2a2) = 2A{ cos^-1 (0.66)}/ π.
Therefore, the total percentage of the period does the mass lie in these regions = 100 × {2a1 + ( A - 2a2) }/A = 2 { cos^-1 (0.66)}/ π × 100 = 54%.
Thus, the total percentage of the period does the mass lie in these regions = 54%.