Cargando....
|
Mantén pulsado para arrastrar. |
|||
|
Haz clic aquí para cerrar |
|||
Pregunta 1 Informe
Detalles de la respuesta
Pregunta 2 Informe
Pregunta 3 Informe
Pregunta 4 Informe
Pregunta 5 Informe
Pregunta 6 Informe
Pregunta 7 Informe
Pregunta 8 Informe
Pregunta 9 Informe
Pregunta 10 Informe
Pregunta 11 Informe
Detalles de la respuesta
Pregunta 12 Informe
Pregunta 13 Informe
Pregunta 14 Informe
Detalles de la respuesta
Pregunta 15 Informe
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
Detalles de la respuesta
Pregunta 17 Informe
Detalles de la respuesta
Pregunta 18 Informe
Detalles de la respuesta
Pregunta 19 Informe
Pregunta 20 Informe
Pregunta 21 Informe
Pregunta 22 Informe
Pregunta 23 Informe
Detalles de la respuesta
Pregunta 24 Informe
Pregunta 25 Informe
Pregunta 26 Informe
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
Pregunta 27 Informe
Detalles de la respuesta
Pregunta 28 Informe
Pregunta 29 Informe
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.
Pregunta 30 Informe
A body is dropped from a certain height above the ground level, falls with uniform
Pregunta 31 Informe
The diagram above represent displacement D, versus time, t graph of a progressive wave. Deduce the frequency of the wave
Detalles de la respuesta
Pregunta 32 Informe
Pregunta 33 Informe
Pregunta 34 Informe
Pregunta 35 Informe
Pregunta 36 Informe
Pregunta 37 Informe
Pregunta 38 Informe
Pregunta 39 Informe
Pregunta 40 Informe
Pregunta 41 Informe
Detalles de la respuesta
Pregunta 42 Informe
Pregunta 43 Informe
Pregunta 44 Informe
Pregunta 45 Informe
Pregunta 46 Informe
Pregunta 47 Informe
Pregunta 48 Informe
When compared with water, mercury is preferred as a thermometric liquid because
Pregunta 50 Informe
Detalles de la respuesta
Pregunta 51 Informe
¿Te gustaría proceder con esta acción?