This question assesses understanding of hydrostatic pressure in fluids, which increases with depth. In a fluid at rest, pressure at a depth h is P = ρgh + P_atm, where ρ is density, g is gravity, and h is depth, showing pressure grows linearly with depth. Point A at 0.20 m has less pressure than Point B at 0.50 m due to the greater overlying fluid weight at B. Thus, Point B experiences greater water pressure. Distractor C suggests equal pressure because the container is sealed, but sealing does not affect the depth dependence. A useful strategy is to compare depths directly, as pressure differences depend solely on Δh in the same fluid.

This question tests pressure due to contact force, defined as force per unit area. Pressure is P = F/A, where F is the perpendicular force on the surface. The 400 N crate is supported by four feet, so each foot bears 100 N, and with area 1.0 × $10^{-3}$ m², the pressure per foot is 100 / 0.001 = 1.0 × $10^5$ Pa. This calculation assumes even weight distribution across the feet. Distractor B (4.0 × $10^5$ Pa) might come from using the total force instead of per foot. When analyzing multi-support contacts, divide the total force equally among supports before applying P = F/A.

This question tests hydrostatic pressure, emphasizing its increase with fluid depth. Hydrostatic pressure is P = ρgh + P0, directly proportional to depth h below the surface. Point C at 1.0 m has lower pressure than Point D at 3.0 m, where the pressure is three times greater due to triple the depth. Therefore, pressure is greater at D because of the increased depth. Choice B is a distractor claiming equal pressure since both are in the same lake, ignoring depth's role. For such problems, always focus on the depth from the surface as the determining factor for pressure comparisons.

This question assesses hydrostatic pressure, focusing on its dependence on vertical depth rather than horizontal position. In fluids, pressure increases with depth h as P = ρgh + P0, but is independent of horizontal distance. Points G and H are both at 2.0 m depth, so they have equal pressure despite the 10 m horizontal separation. The pressures are the same because only vertical depth matters. Choice B is a distractor that incorrectly claims pressure increases with horizontal distance, which is not true in static fluids. A transferable strategy is to ignore horizontal positions and base comparisons solely on vertical depths from the surface.

This question tests understanding of the inverse relationship between pressure and area. Pressure is defined as P = F/A, so if force remains constant at 200 N while area doubles, pressure becomes half of its original value. Mathematically, if initial pressure is P₁ = F/A, then new pressure is P₂ = F/(2A) = (F/A)/2 = P₁/2. Choice B incorrectly suggests pressure remains constant, ignoring the role of area in the pressure equation. To solve pressure problems with changing conditions, apply P = F/A systematically and note which variables change.

This question tests understanding of pressure variation with depth in fluids. In any static fluid, pressure increases with depth following P = P₀ + ρgh. Since point M is 0.20 m deeper than point N, M experiences greater pressure due to the additional weight of water above it. The pressure difference equals ρg(0.20 m), where ρ is water density and g is gravitational acceleration. Choice B incorrectly suggests pressure decreases with depth, which violates basic fluid statics principles. When analyzing fluid pressure, remember that pressure always increases as you go deeper, regardless of the fluid type or container.

This question tests understanding that pressure depends only on depth in a static fluid. In any static fluid, pressure at a given depth is P = P₀ + ρgh, which depends only on the vertical depth h below the surface, not on horizontal position. Since both points A and B are at the same 0.30 m depth, they experience identical pressure regardless of their horizontal separation or position relative to container walls. This principle allows pressure to be transmitted equally throughout a fluid at the same depth. Choice A incorrectly suggests that being "directly under" matters, confusing vertical depth with horizontal position. When analyzing fluid pressure, only the vertical depth below the surface determines the pressure value.

This question tests understanding of pressure as force per unit area. With the same 40 N weight but different contact areas, pressure varies inversely with area according to P = F/A. For area A₁ = 0.020 m², pressure is P₁ = 40/0.020 = 2000 Pa. For smaller area A₂ = 0.010 m², pressure is P₂ = 40/0.010 = 4000 Pa, which is greater. Choice C incorrectly assumes equal pressures because weight is constant, missing that pressure depends on both force and area. To compare pressures with constant force, remember that smaller contact area always produces greater pressure.

This question tests understanding of pressure variation with depth in fluids. In a static fluid, pressure increases linearly with depth according to P = P₀ + ρgh, where ρ is fluid density, g is gravitational acceleration, and h is depth below the surface. Since point Q is 2.0 m deeper than point P, the pressure at Q must be greater than at P due to the additional weight of water above it. Choice B incorrectly suggests equal pressures, confusing the fact that pressure acts in all directions with pressure magnitude. When comparing pressures in fluids, always consider the depth difference and remember that deeper points have higher pressure.

This question tests understanding of pressure as force per unit area. Pressure is calculated as P = F/A, where F is specifically the normal force (perpendicular to the surface) and A is the contact area. Since the normal force is the same in both cases but Case B has half the contact area (0.25 m²) compared to Case A (0.50 m²), Case B produces twice the pressure. The horizontal friction force from pulling does not contribute to vertical pressure. Choice D incorrectly suggests friction affects pressure, when only the perpendicular normal force matters. For pressure calculations, always identify which force component is perpendicular to the surface.