🔑:Given that the vapor pressure of the substance at 20°C is 1.5 atm and the atmospheric pressure outside the container is 1 atm, we can infer what will happen to the substance inside the container based on the principles of vapor pressure, volatility, and evaporation.## Step 1: Understanding Vapor PressureVapor pressure is the pressure exerted by the vapor of a substance in equilibrium with its liquid phase at a given temperature. It is a measure of the tendency of the molecules to escape from the surface of the liquid and turn into vapor.## Step 2: Comparing Vapor Pressure to Atmospheric PressureSince the vapor pressure of the substance (1.5 atm) is greater than the atmospheric pressure outside the container (1 atm), it indicates that the substance has a high tendency to evaporate. The difference between the vapor pressure and the external pressure suggests that the substance is volatile at the given temperature.## Step 3: Determining the OutcomeWhen the vapor pressure of a substance exceeds the external pressure, the substance will evaporate rapidly. However, since the container is closed, the evaporation will continue until the pressure inside the container equals the vapor pressure of the substance. This means that the air space above the liquid will fill with vapor until equilibrium is reached, at which point the rate of evaporation will equal the rate of condensation.## Step 4: ConclusionGiven the closed system, the substance will not continuously evaporate to the outside but will instead reach an equilibrium state where the vapor pressure inside the container matches the substance's vapor pressure at 20°C. The substance inside the container will partially evaporate, creating a pressure of 1.5 atm inside the container, which is higher than the external atmospheric pressure. This higher internal pressure is due to the vapor of the substance, indicating that the liquid and vapor phases of the substance are in equilibrium within the closed container.The final answer is: The substance will partially evaporate until the pressure inside the container reaches 1.5 atm, at which point the rates of evaporation and condensation will be equal, and equilibrium will be established.

🔑:To solve this problem, we'll break it down into steps focusing on the physics involved, specifically the equations of motion under gravity and the relationship between the angle of impact and the initial velocity components.## Step 1: Identify the knowns and unknowns- Known: Height of the cliff (h) = 20 m- Known: Angle of impact with the ground = 45 degrees- Unknown: Initial speed (v₀) of the ball- Assumption: Neglect air resistance## Step 2: Determine the time it takes for the ball to hit the groundSince the ball is thrown horizontally, the vertical component of its velocity is initially 0 m/s. It accelerates downward due to gravity (g = 9.81 m/s²). The time it takes to hit the ground can be found using the equation for free fall: h = (1/2)gt², where h is the height of the fall.Rearranging for t gives: t = √(2h/g)## Step 3: Calculate the timeSubstitute the known values into the equation: t = √(2*20/9.81) = √(40/9.81) = √4.07 ≈ 2.02 seconds## Step 4: Find the vertical component of velocity at impactThe vertical component of velocity (v_y) at the moment of impact can be found using the equation v_y = gt, since the initial vertical velocity is 0.v_y = 9.81 * 2.02 ≈ 19.84 m/s## Step 5: Use the angle of impact to find the horizontal component of velocityAt impact, the ball strikes the ground at a 45-degree angle. This means the vertical and horizontal components of its velocity are equal in magnitude because tan(45°) = v_y / v_x = 1, implying v_x = v_y.## Step 6: Calculate the initial speedThe initial speed (v₀) is the horizontal component of velocity since the ball was thrown horizontally. Given v_x = v_y, and knowing v_y at impact, we can say v₀ = v_x = 19.84 m/s.However, this step simplifies to recognizing that since v_x = v_y at the point of impact due to the 45-degree angle, and we've calculated v_y, we directly have the magnitude of the velocity components. The initial speed (v₀) is thus equal to the horizontal component of velocity, which we've determined to be equal to the vertical component at impact due to the angle.The final answer is: boxed{19.84}

🔑:## Step 1: Define the variables and the goalThe boy can row at a speed of 2 m/s in still water, and the river is flowing at 1 m/s. We need to find the angle (theta) at which he should point the bow of his boat to cross the river in the shortest possible time.## Step 2: Understand the components of the boy's velocityWhen the boy rows at an angle theta to the riverbank, his velocity can be resolved into two components: one perpendicular to the riverbank (which we'll call the "cross-river" component) and one parallel to the riverbank (which we'll call the "along-river" component).## Step 3: Calculate the cross-river component of the boy's velocityThe cross-river component of the boy's velocity is given by 2 * sin(theta), since this is the component of his rowing velocity that is directed across the river.## Step 4: Calculate the along-river component of the boy's velocityThe along-river component of the boy's velocity is given by 2 * cos(theta) - 1, since this is the component of his rowing velocity that is directed along the river, minus the speed of the river itself.## Step 5: Determine the condition for the shortest crossing timeTo achieve the shortest crossing time, the boy should minimize his along-river velocity component (since a larger along-river velocity would increase his travel time), while maximizing his cross-river velocity component.## Step 6: Apply the condition for the shortest crossing timeThe boy should row at an angle that maximizes his cross-river velocity component. However, we also need to consider the effect of the river's flow. The optimal angle occurs when the along-river velocity component is zero, i.e., when 2 * cos(theta) - 1 = 0.## Step 7: Solve for thetaRearranging the equation from Step 6, we get 2 * cos(theta) = 1. Solving for theta, we find cos(theta) = 1/2, which implies theta = arccos(1/2).## Step 8: Calculate the value of thetaThe arccos(1/2) is equal to 60 degrees.The final answer is: boxed{60}

🔑:## Step 1: Recall the mass-energy equivalence formulaThe formula that relates mass (m) and energy (E) is given by E = mc^2, where c is the speed of light in a vacuum (approximately 3 * 10^8 m/s).## Step 2: Calculate the mass equivalent of the released energyUsing the formula E = mc^2, we can find the mass equivalent of the released energy. Given E = 10^6 J, we rearrange the formula to solve for m: m = E / c^2.## Step 3: Perform the calculation for the mass equivalentSubstitute the given values into the formula: m = 10^6 J / (3 * 10^8 m/s)^2 = 10^6 J / (9 * 10^16 m^2/s^2) = 10^6 / (9 * 10^16) kg = 1.11 * 10^-11 kg.## Step 4: Calculate the final mass of the boxThe final mass of the box is the initial mass minus the mass equivalent of the energy released, since the energy released corresponds to a loss of mass according to the mass-energy equivalence principle. Therefore, final mass = initial mass - mass equivalent of released energy.## Step 5: Perform the calculation for the final massGiven the initial mass is 10 kg, and the mass equivalent of the released energy is 1.11 * 10^-11 kg, the final mass = 10 kg - 1.11 * 10^-11 kg.The final answer is: boxed{9.99999999889}