WAEC SSCE - Physics - 2004 (Objective)
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When compared with water, mercury is preferred as a thermometric liquid because
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A body is dropped from a certain height above the ground level, falls with uniform
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In an experiment using a converging lens to produce real images on a screen, the linear magnification m, is plotted against the image distance v, as illustrated in the diagram below. The distance OP represents the
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The diagram above represent displacement D, versus time, t graph of a progressive wave. Deduce the frequency of the wave
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Weight of object in air (W_air): 60.0 N
Weight of object in liquid X (W_X): 48.2 N
Weight of object in water (W_water): 44.9 N
The loss of weight in a fluid is equal to the buoyant force, which is equal to the weight of the displaced fluid.
Buoyant force in water: 𝐹buoyant, water=𝑊air−𝑊water=60.0 N−44.9 N=15.1 NFbuoyant, water=Wair−Wwater=60.0N−44.9N=15.1N
Since the buoyant force is equal to the weight of the water displaced and the density of water (ρ_water) is 1 g/cm³ (or 1000 kg/m³), the volume of the object (V) can be calculated using:
𝐹buoyant, water=𝜌water⋅𝑉⋅𝑔Fbuoyant, water=ρwater⋅V⋅g 15.1 N=1000 kg/m3⋅𝑉⋅9.8 m/s215.1N=1000kg/m3⋅V⋅9.8m/s2 𝑉=15.1 N1000 kg/m3⋅9.8 m/s2V=1000kg/m3⋅9.8m/s215.1N 𝑉≈1.54×10−3 m3V≈1.54×10−3m3
Buoyant force in liquid X:
𝐹buoyant, X=𝑊air−𝑊X=60.0 N−48.2 N=11.8 NFbuoyant, X=Wair−WX=60.0N−48.2N=11.8N
Using the same volume V (since the object's volume doesn't change), the density of liquid X (ρ_X) can be calculated:
𝐹buoyant, X=𝜌X⋅𝑉⋅𝑔Fbuoyant, X=ρX⋅V⋅g 11.8 N=𝜌X⋅1.54×10−3 m3⋅9.8 m/s211.8N=ρX⋅1.54×10−3m3⋅9.8m/s2 𝜌X=11.8 N1.54×10−3 m3⋅9.8 m/s2ρX=1.54×10−3m3⋅9.8m/s211.8N 𝜌X≈786.1 kg/m3ρX≈786.1kg/m3
Relative density of liquid X: The relative density (specific gravity) is the ratio of the density of liquid X to the density of water: Relative density=𝜌X𝜌water=786.1 kg/m31000 kg/m3≈0.786Relative density=ρwaterρX=1000kg/m3786.1kg/m3≈0.786
Therefore, the relative density of liquid X is approximately 0.786.
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A loaded spring performs simple harmonic motion with an amplitude of 5cm. If the maximum acceleration of the load is 20cm s-1, calculate the angular frequency of the motion
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