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As shown in the figure, PQ is a thin metal sheet in a uniform magnetic field, and its plane is parallel to the direction of the magnetic field. A charged particle shoots out from a certain point at a velocity V perpendicular to PQ, and the kinetic energy is e. the trajectory of the charged particle after shooting is shown in the figure. The ratio of its trajectory radius on both sides of the metal sheet is 10:9. If the resistance and electric quantity of the particle are constant in the process of passing through the plate, it can be seen as follows, Then () A. Charged particles must be positively charged. B. the velocity of charged particles decreases by 1102emc every time they pass through the metal sheet. The kinetic energy of charged particles decreases by 0.19ed every time they pass through the metal sheet. The charged particles are trapped in the metal sheet after they pass through five times
According to the left-handed rule, we know that the particle is positively charged, so a is correct; according to the centripetal force provided by Lorentz force: QVB = mv2rv = qbrm, the ratio of the radius of its trajectory on both sides of the metal sheet is 10:9, so: v1v0 = r1r0 = 910, that is: V1 = 0.9v0 △ v = 0.1v0 = 0.12em, so B is correct; the kinetic energy of the particle: EK = 12mv2, the kinetic energy lost after the particle passes through the metal sheet each time: △ EK = 12mv02-12mv12 = 12mv02 (1-0.92) ）=The number of times of particle passing through the metal sheet: n = e0.19e = 5.26, i.e. α particle is trapped in the metal sheet after passing through the metal sheet for 5 times; so CD is correct; so select ABCD
A monument weighs 106n and stands on a cornerstone with a height of 1m and a density of 2 × 103kg / m3. If the pressure on the ground can not exceed 7 × 104pa, at least what is the bottom area of the cornerstone? （g=10N/Kg）
∵ P = FS, ∵ pressure f = PS, the total gravity of the base and monument is equal to the maximum pressure on the ground, let the bottom area of the cornerstone be s, (G0 + ρ base SH) g = P, the maximum s, that is: 106n + 2 × 103kg / m3 × s × 1m × 10N / kg = 7 × 104pa × s, the solution is: S = 20m2
A rectangular wood block with density of ρ = 0.6 × 103 kg / m3, length a = 0.4m, width b = 0.2m, height C = 0.1M. (1) when the wood block floats on the water, what is the volume of the part above the water? (2) If we use the method of hollowing out to make this block into a small "boat", so that it can carry 5.9 kg of sand without sinking, how large should the volume of the excavated part of the block be? (density of water ρ = 1 × 103 kg / m3, g = 10N / kg)
(1) The volume of wood block is: v = ABC = 0.4m × 0.2m × 0.1M = 0.008m3. The gravity of wood block is: G wood = mg = ρ wood GV = 0.6 × 103kg / m3 × 10N / kg × 0.008m3 = 48n. Because the wood block floats, f floating = g wood = 48n is discharged according to f floating = ρ water GV, V drainage = f floating ρ water g = 48n1.0 × 103kg / m3 × 10N / kg = 4.8 × 10-3m3. The volume of the exposed part is v exposed = 8 × 10-3m3-4.8 × 10-3m3 = 3.2 × 10-3m3 (2) when the wood block is completely submerged, the buoyancy is f ′ = ρ water GV = 1.0 × 103kg / m3 × 10N / kg × 0.008m3 = 80N, the gravity of the remaining wood block and sand is g total = f floating = 80N, so the mass of the remaining wood block is m residual = g total G-M sand = 80n10n / kg-5.9kg = 2.1kg, the volume of the remaining wood block is v residual = m residual ρ wood = 2.1kg 0.6 × 103kg / m3 = 3.5 × 10-3m3, so the volume of the excavated wood block is v residual = m residual ρ wood = 2.1kg 0.6 × 103kg / m3 = 3.5 × 10-3m3 : V excavation = 0.008m3-2.1kg 0.6 × 103kg / m3 = 4.5 × 10-3m3 answer: the volume of the excavated part is at least 4.5 × 10-3m3
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