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The diagram above illustrates a waveform. Which of the point on the waveform are in phase?
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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\(^{-2}\)]
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The rising of a liquid in an open and ended glass tube of narrow bore is
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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\(^{-1}\)]
Bayanin Amsa
Tambaya 10 Rahoto
Solid friction, like viscosity is
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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
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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
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Bayanin Amsa
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Bayanin Amsa
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Bayanin Amsa
Tambaya 21 Rahoto
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
Bayanin Amsa
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Bayanin Amsa
Tambaya 25 Rahoto
The total area under a force-velocity graph represents
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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?
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Tambaya 30 Rahoto
A string under tension produces a note of frequency 14Hz. Determine the frequency when the tension is quadrupled.
Tambaya 31 Rahoto
Which of the following units is equivalent to the watt?
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Tambaya 33 Rahoto
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.
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When salt is dissolved in water, the freezing point of the water
Tambaya 36 Rahoto
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
Tambaya 37 Rahoto
A potential difference of 12V is applied across the ends of a 6\(\Omega\) resistor for 10 minutes. Determine the quantity of heat generated
Tambaya 38 Rahoto
A diverging lens of focal length 30cm produces an image 20cm from the lens. Determine the object distance
Tambaya 39 Rahoto
The refractive index of a material is 1.5. Calculate the critical angle at the glass-air interface
Tambaya 40 Rahoto
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
Tambaya 41 Rahoto
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
Tambaya 42 Rahoto
Which of the following statements about pressure in a liquid is correct?
Tambaya 43 Rahoto
Images formed by a convex mirror are always
Tambaya 45 Rahoto
Which of the following phenomena supports the theory that waves have a particle nature?
Tambaya 46 Rahoto
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
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