4.3. Density and pressure
A subsection of Physics, 9702, through 4. Forces, density and pressure
Listing 10 of 120 questions
The drag force FD acting on a sphere falling through a liquid is given by FD = 6πηr v where r is the radius of the sphere, v is the speed of the sphere in the liquid and η is a property of the liquid called the viscosity. Show that the SI base units of viscosity are kg m–1 s–1. The sphere has a radius of 3.0 cm and is falling vertically downwards at a terminal velocity of 2.0 m s–1 through the liquid. The drag force acting on the sphere is 0.096 N. Calculate the viscosity of the liquid. viscosity = kg m–1 s–1 The sphere is shown in . sphere liquid On , draw and label arrows to represent the directions of the three forces acting on the sphere as it falls at terminal velocity through the liquid. The density of the liquid is 920 kg m–3. Show that the upthrust acting on the sphere is 1.0 N. Calculate the mass of the sphere. mass = kg
9702_s24_qp_23
THEORY
2024
Paper 2, Variant 3
View
A cylinder is suspended from the end of a string. The cylinder is stationary in water with the axis of the cylinder vertical, as shown in . h string cylinder weight 0.84 N water density 1.0 × 103 kg m–3 0.031 m (not to scale) The cylinder has weight 0.84 N, height h and a circular cross-section of diameter 0.031 m. The density of the water is 1.0 × 103 kg m−3. The difference between the pressures on the top and bottom faces of the cylinder is 520 Pa. Calculate the height h of the cylinder. h = m Show that the upthrust acting on the cylinder is 0.39 N. Calculate the tension T in the string. T = N The string is now used to move the cylinder in vertically upwards through the water. The variation with time t of the velocity v of the cylinder is shown in . 12.5 10.0 7.5 5.0 2.5 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 v / cm s–1 t / s Use to determine the acceleration of the cylinder at time t = 2.0 s. acceleration = m s−2 The top face of the cylinder is at a depth of 0.32 m below the surface of the water at time t = 0. Use to determine the depth of the top face below the surface of the water at time t = 4.0 s. depth = m The cylinder in is released from the string at time t = 4.0 s. The cylinder falls, from rest, vertically downwards through the water. Assume that the upthrust acting on the cylinder remains constant as it falls. State the name of the force that acts on the cylinder when it is moving and does not act on the cylinder when it is stationary. State and explain the variation, if any, of the acceleration of the cylinder as it falls downwards through the water.
9702_w20_qp_22
THEORY
2020
Paper 2, Variant 2
View
Complete by putting a tick (3) in the appropriate column to indicate whether the listed quantities are scalars or vectors. quantity scalar vector acceleration force kinetic energy momentum power work A floating sphere is attached by a cable to the bottom of a river, as shown in . water surface solid sphere direction of flow of water cable river bed 75° The sphere is in equilibrium, with the cable at an angle of 75° to the horizontal. Assume that the force on the sphere due to the water flow is in the horizontal direction. The radius of the sphere is 23 cm. The sphere is solid and is made from a material of density 82 kg m–3. Show that the weight of the sphere is 41 N. The tension in the cable is 290 N. Determine the upthrust acting on the sphere. upthrust = N Explain the origin of the upthrust acting on the sphere.
9702_m17_qp_22
THEORY
2017
Paper 2, Variant 2
View
Questions Discovered
120