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Ajụjụ 1 Ripọtì
Ajụjụ 2 Ripọtì
The total area under a force-velocity graph represents
Ajụjụ 3 Ripọtì
The outlet of a bicycle pump is used to inflate a football as illustrated in the diagram above. Why does the pressure of the air inside the pump increases as the pump handle is slowly pushed downward at constant temperature?
Ajụjụ 5 Ripọtì
Ajụjụ 6 Ripọtì
The rising of a liquid in an open and ended glass tube of narrow bore is
Ajụjụ 7 Ripọtì
Which of the following statements about pressure in a liquid is correct?
Ajụjụ 8 Ripọtì
Ajụjụ 9 Ripọtì
A charge of 2.0 x 10\(^{-5}\) C experiences a force of 80N in a uniform electric field. Calculate the magnitude of the electric field intensity
Ajụjụ 10 Ripọtì
Akọwa Nkọwa
Ajụjụ 11 Ripọtì
Ajụjụ 12 Ripọtì
Ice of mass 10g at 5°C was completely converted to water at 0°C, calculate the quantity of heat used. [specific heat capacity of ice = 2.1 Jg\(^{-1}\)K\(^{-1}\), specific latent heat of fusion of ice = 336 Jg\(^{-1}\)]
Akọwa Nkọwa
Ajụjụ 13 Ripọtì
Ajụjụ 14 Ripọtì
Which of the following units is equivalent to the watt?
Ajụjụ 15 Ripọtì
A body of mass 2kg is released from rest on a plane inclined at an angle of 60° to the horizontal. Calculate the acceleration of the body down the plane.[g = 10ms\(^{-2}\)]
Ajụjụ 16 Ripọtì
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Ajụjụ 18 Ripọtì
Ajụjụ 19 Ripọtì
When salt is dissolved in water, the freezing point of the water
Ajụjụ 20 Ripọtì
To determine the potential energy of the body at the maximum height of its motion, we need to follow these steps:
Calculate the speed of projection (initial velocity): From the previous solution, we determined that the speed of projection u is 6 m/s.
Determine the maximum height: Use the kinematic equation for the vertical motion to find the maximum height:
v2=u2−2ghAt the maximum height, the final velocity v is 0, so:
0=u2−2gh⟹h=2gu2Substituting u=6 m/s and g=10 m/s2:
h=2×10 m/s2(6 m/s)2=20 m/s236 m2/s2=1.8 mCalculate the potential energy (PE): The potential energy at the maximum height is given by:
PE=mghConvert the mass from grams to kilograms:
m=20 g=0.02 kgNow substitute m=0.02 kg, g=10 m/s2, and h=1.8 m:
PE=0.02 kg×10 m/s2×1.8 m=0.36 JTherefore, the potential energy of the body at the maximum height of its motion is 0.36 J
Ajụjụ 21 Ripọtì
A block of wood of mass 5kg is pulled on a platform by a force of 40N as illustrated in the diagram above. If the frictional force, F experienced by the block is 12N, calculate the magnitude of the acceleration of the acceleration of the block
Ajụjụ 22 Ripọtì
Ajụjụ 23 Ripọtì
Ajụjụ 24 Ripọtì
An electric heater has a resistance of 50\(\Omega\). When it is immersed in water and connected to mains source, it draws a current of 4.0A. Calculate the heat gained by the water if the heater is switched on for 2 minutes, assuming no heat losses to the surroundings
Ajụjụ 25 Ripọtì
Ajụjụ 26 Ripọtì
A string under tension produces a note of frequency 14Hz. Determine the frequency when the tension is quadrupled.
Ajụjụ 27 Ripọtì
Ajụjụ 28 Ripọtì
A diverging lens of focal length 30cm produces an image 20cm from the lens. Determine the object distance
Ajụjụ 29 Ripọtì
An electric circuit is connected as illustrated above. Determine the equivalent e.m.f and current flowing through the circuit respectively, neglecting the internal resistance of the cells
Akọwa Nkọwa
Ajụjụ 30 Ripọtì
The refractive index of a material is 1.5. Calculate the critical angle at the glass-air interface
Ajụjụ 31 Ripọtì
A body of mass 20g projected vertically upwards in vacuum returns to the point of projection after 1.2s. [g = 10ms\(^{-2}\)]. Calculate the speed of projection
To determine the speed of projection of a body projected vertically upwards in a vacuum, we can use the principles of kinematics and the given information. Here are the steps:
Understand the total time of flight: The total time taken for the body to go up and come back down is 1.2 seconds.
Determine the time to reach the maximum height: Since the time to go up is equal to the time to come down, the time to reach the maximum height is half of the total time of flight.
tup=21.2 s=0.6 sUse the kinematic equation for velocity: The velocity at the maximum height is 0 (since the body momentarily stops before falling back down). We can use the equation:
v=u−gtwhere v is the final velocity (0 m/s at the maximum height), u is the initial velocity (speed of projection), g is the acceleration due to gravity (10 m/s²), and t is the time taken to reach the maximum height (0.6 s).
Setting v=0:
0=u−g⋅tSolve for u:
u=g⋅t=10 m/s2⋅0.6 s=6 m/sTherefore, the speed of projection is
6.0 m/s
Ajụjụ 32 Ripọtì
Akọwa Nkọwa
Ajụjụ 33 Ripọtì
A potential difference of 12V is applied across the ends of a 6\(\Omega\) resistor for 10 minutes. Determine the quantity of heat generated
Akọwa Nkọwa
Ajụjụ 34 Ripọtì
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
Images formed by a convex mirror are always
Ajụjụ 36 Ripọtì
Ajụjụ 37 Ripọtì
Ajụjụ 38 Ripọtì
Four co-planar forces of magnitudes 10N, 17N, 6N and 20N act at point O as shown in the above diagram. Determine the magnitude of the resultant force
Akọwa Nkọwa
Ajụjụ 39 Ripọtì
The diagram above illustrates a waveform. Which of the point on the waveform are in phase?
Akọwa Nkọwa
Ajụjụ 40 Ripọtì
Akọwa Nkọwa
Ajụjụ 42 Ripọtì
Solid friction, like viscosity is
Ajụjụ 44 Ripọtì
Which of the following phenomena supports the theory that waves have a particle nature?
Ajụjụ 45 Ripọtì
The equation of a certain progressive transverse wave is y = 2 sin 2 \(\pi ( \frac{t}{0.01} - \frac{x}{30})\), where x and y are in cm and t in seconds. Calculate the period of the wave.
Ajụjụ 46 Ripọtì
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