answer:The work done by a force, denoted as ( W ), can be calculated using the formula: [ W = Fdcos(theta) ] Here, ( F ) represents the applied force, ( d ) is the displacement, and ( theta ) is the angle between the force and displacement vectors. 1. Positive Work ( ( W > 0 ) ): When ( 0^circ leqslant theta < 90^circ ) and ( 270^circ < theta leqslant 360^circ ), the work done is positive. This occurs when the force and displacement are in the same or opposite directions. (a) For instance, if someone pushes a box along a flat surface in the direction of motion, the angle ( theta ) is ( 0^circ ), and the work done is positive because the force and displacement align. 2. Zero Work ( ( W = 0 ) ): At ( theta = 90^circ ) and ( theta = 270^circ ), the work is zero. This means the force and displacement are perpendicular. 3. Negative Work ( ( W < 0 ) ): When ( 90^circ < theta < 270^circ ), the work done is negative, occurring when the force and displacement are in opposite directions. (b) An example of negative work is the gravitational force acting on an object rising vertically against gravity. In this case, ( theta = 180^circ ) since gravity pulls downward while the object moves upward. Thus, the work done by gravity is negative. These examples demonstrate the relationship between the angle ( theta ) and the sign of the work done, reflecting the interaction between the force and the displacement.

answer:To determine the decay model for the substance, we use the exponential decay formula {eq}A = Pe^{-kt} {/eq}, where A is the amount remaining, P is the initial amount, k is the decay constant, and t is time. Given that 95% remains after one year, we have {eq}0.95P = Pe^{-k(1)} {/eq}. This simplifies to {eq}0.95 = e^{-k} {/eq}. To find k, we take the natural logarithm of both sides: {eq}ln(0.95) = -k {/eq} Hence, the decay function is {eq}A = Pe^{-ln(0.95) t} {/eq}. Since k is negative (as the logarithm of a number less than one is negative), the decay constant incorporates this sign. a) The half-life is the time required for half of the original amount to decay, i.e., {eq}0.5P = Pe^{-ln(0.95) t} {/eq}. Solving for t: {eq}0.5 = e^{-ln(0.95) t} {/eq} {eq}ln(0.5) = -ln(0.95) t {/eq} {eq}t = frac{ln(0.5)}{-ln(0.95)} {/eq} {eq}t approx 13.51 text{ years} {/eq} b) To find the time for the sample to decay to 25% of its original amount, we set {eq}0.25P = Pe^{-ln(0.95) t} {/eq}: {eq}0.25 = e^{-ln(0.95) t} {/eq} {eq}ln(0.25) = -ln(0.95) t {/eq} {eq}t = frac{ln(0.25)}{-ln(0.95)} {/eq} {eq}t approx 27.03 text{ years} {/eq} Therefore, the half-life of the substance is approximately 13.51 years, and it will take about 27.03 years to decay to 25% of its original quantity.

answer:Speed of longitudinal waves in a long steel rod: {eq}v_{steel}=sqrt{frac{Y_{steel}}{rho_{steel}}}=sqrt{frac{200 times 10^9 rm Nm^{-2}}{7.8 times 10^3 rm kgm^{-3}}}=5.09 times 10^3 rm ms^{-1}{/eq} Speed of longitudinal waves in a long aluminum rod: {eq}v_{aluminum}=sqrt{frac{Y_{aluminum}}{rho_{aluminum}}}=sqrt{frac{70 times 10^9 rm Nm^{-2}}{2.7 times 10^3 rm kgm^{-3}}}=5.11 times 10^3 rm ms^{-1}{/eq}

answer:The Euler totient function, phi(n), counts the number of positive integers less than n that are coprime to n. To compute phi(38029), we note that 38029 is a prime number. For a prime number p, phi(p) = p - 1. Therefore, phi(38029) = 38029 - 1 = 38028.