answer:The circulation of a vector field around a closed curve can be found using Stokes' Theorem, which states: {eq}oint_C vec F cdot dvec r = iint_S text{curl}(vec F) cdot hat n , dS {/eq} Here, {eq}C {/eq} is the triangle with vertices (4,0,0), (4,0,4), and (4,6,4), and {eq}S {/eq} is the region enclosed by {eq}C {/eq} in the plane {eq}x = 4 {/eq}. The unit normal {eq}hat n {/eq} for this surface, considering a clockwise orientation for {eq}C {/eq}, points in the negative x-direction, i.e., {eq}hat n = -hat i {/eq}. The curl of the vector field {eq}vec F {/eq} is: {eq}text{curl}(vec F) = begin{vmatrix} hat i & hat j & hat k frac{partial}{partial x} & frac{partial}{partial y} & frac{partial}{partial z} 7y & 7z & 5x end{vmatrix} = langle -7, -5, -7 rangle {/eq} So, the circulation is: {eq}begin{align*} oint_C vec F cdot dvec r &= iint_S langle -7, -5, -7 rangle cdot (-hat i) , dA &= 7 iint_S dA &= 7 times text{Area of } S &= 7 times left(frac{1}{2} times text{base} times text{height}right) &= 7 times left(frac{1}{2} times 6 times 4right) therefore text{Circulation} &= 84 end{align*} {/eq} The base and height of the triangle are derived from the vertices: {eq}text{base} = 6 , (text{difference in } y) & text{height} = 4 , (text{difference in } z) {/eq}.

answer:Given data: The length of the first train is {eq}{L_1} = 120;{rm{m}}.{/eq} The velocity of the first train is {eq}{v_1} = 120;{rm{kph}} times dfrac{{0.278;{rm{m/s}}}}{{1;{rm{kph}}}} = 33.36;{rm{m/s}}.{/eq} The length of the second train is {eq}{L_2} = 180;{rm{m}}.{/eq} The velocity of the second train is {eq}{v_2} = 200;{rm{kph}} times dfrac{{0.278;{rm{m/s}}}}{{1;{rm{kph}}}} = 55.6;{rm{m/s}}.{/eq} The distance between the front ends of the trains is {eq}d = 1000;{rm{m}}.{/eq} Since the trains are moving in opposite directions, their relative velocity is the sum of their individual velocities. {eq}{v_{rel}} = {v_1} + {v_2} = 33.36;{rm{m/s}} + 55.6;{rm{m/s}} = 88.96;{rm{m/s}}.{/eq} The time it takes for the trains to completely pass each other is the time it takes for them to cover the sum of their lengths plus the distance between their front ends. {eq}t = dfrac{{d + {L_1} + {L_2}}}{{{v_{rel}}}}{/eq} Substitute the values in the above equation. {eq}begin{align*} t &= dfrac{{1000;{rm{m}} + 120;{rm{m}} + 180;{rm{m}}}}{{88.96;{rm{m/s}}}} t &= 14.66;{rm{s}} end{align*} {/eq} Thus, the time it takes for the trains to completely pass each other is {eq}14.66;{rm{s}}. {/eq}

answer:Given: Spring constant, k = 1400 N/m Mass of the diver, m = 100 kg Density of the diver, ρ = 0.65 g/cm³ = 650 kg/m³ Change in length, ΔL The diver is in equilibrium in the water, so the net force acting on him is zero. There are three forces acting on the diver: 1. Weight (W) force, downwards 2. Spring force, downwards 3. Buoyant force (F), upwards Since the diver is in equilibrium, we can write: W + kΔL = F ρgV + kΔL = ρ_w gV where ρ is the density of the diver, g is the acceleration due to gravity, V is the volume of the diver, ρ_w is the density of water, and V is the volume of the diver submerged in water. We can simplify the equation as: mg + kΔL = ρ_w gV Substituting the given values, we get: (100 kg)(9.81 m/s²) + (1400 N/m)ΔL = (1000 kg/m³)(9.81 m/s²)(100 kg / 650 kg/m³) Simplifying further, we get: 981 N + 1400 N/m ΔL = 151.3 N Solving for ΔL, we get: ΔL = (151.3 N - 981 N) / 1400 N/m ΔL = -0.594 m The negative sign indicates that the spring is compressed, not elongated. Therefore, the elongation of the spring is 0.594 m.

answer:Assuming a two-tailed hypothesis test: {eq}H_0: mu = 45.1[2ex] H_a: mu neq 45.1 {/eq} Case I: - Significance level: {eq}alpha = 0.05 - Sample mean: {eq}bar{x}_1 = 43 - Sample standard deviation: {eq}s_1 = 5 - Sample size: {eq}n = 100 {/eq} Test statistic: {eq}z_1 = dfrac{bar{x}_1 - mu}{dfrac{s_1}{sqrt{n}}} = dfrac{43 - 45.1}{dfrac{5}{sqrt{100}}} = -4.2 {/eq} Decision rule: Reject {eq}H_0 {/eq} if {eq}p-value leq alpha {/eq}. {eq}text{p-value} = P(z > 4.2) + P(z < -4.2) approx 0.00003 {/eq} Since {eq}text{p-value } approx 0.00003 < alpha = 0.05 {/eq}, we reject the null hypothesis. Case II: - Significance level: {eq}alpha = 0.05 - Sample mean: {eq}bar{x}_2 = 43.8 - Sample standard deviation: {eq}s_2 = 8 - Sample size: {eq}n = 100 {/eq} Test statistic: {eq}z_2 = dfrac{bar{x}_2 - mu}{dfrac{s_2}{sqrt{n}}} = dfrac{43.8 - 45.1}{dfrac{8}{sqrt{100}}} = -1.625 {/eq} Decision rule: Reject {eq}H_0 {/eq} if {eq}p-value leq alpha {/eq}. {eq}text{p-value} = P(z > 1.625) + P(z < -1.625) approx 0.10416 {/eq} Since {eq}text{p-value } approx 0.10416 > alpha = 0.05 {/eq}, we fail to reject the null hypothesis. In conclusion, for the first sample, we reject the null hypothesis, indicating a significant difference from the population mean. For the second sample, we do not find sufficient evidence to reject the null hypothesis, suggesting that the sample mean is not significantly different from the population mean at the {eq}alpha = 0.05 {/eq} level.