Ana ebu...
|
Pịa ma Jide iji Dọkpụrụ Ya |
|||
|
Pịa Ebe a ka Imechi |
|||
Ajụjụ 1 Ripọtì
An object projected at an angle to a ground level has a time of flight 4 seconds to move through still air. Calculate the maximum height attained by the object.[g = 10ms\(^{-2}\)]
Given: time of flight T = 4s, g = 10ms\(^{-2}\), H\(_{max}\)
\(\theta\) = ?
T = \(\frac{2u sin\theta}{g}\)
4 = \(\frac{2 \times u sin \theta}{10}\)
40 = 2u sin\(\theta\)
sin\(\theta\) = \(\frac{20}{u}\)
But, H\(_{max}\) = \(\frac{u^2 sin^2 \theta}{2g}\)
H\(_{max}\) = \(\frac{u^2 \times (\frac{20}{u})^2}{2 \times 10}\)
H\(_{max}\) = \(\frac{20 \times 20}{20}\) = 20 metres.(u\(^2\)s cancel out)
Thus H\(_{max}\) attained = 20metres.
Akọwa Nkọwa
Given: time of flight T = 4s, g = 10ms\(^{-2}\), H\(_{max}\)
\(\theta\) = ?
T = \(\frac{2u sin\theta}{g}\)
4 = \(\frac{2 \times u sin \theta}{10}\)
40 = 2u sin\(\theta\)
sin\(\theta\) = \(\frac{20}{u}\)
But, H\(_{max}\) = \(\frac{u^2 sin^2 \theta}{2g}\)
H\(_{max}\) = \(\frac{u^2 \times (\frac{20}{u})^2}{2 \times 10}\)
H\(_{max}\) = \(\frac{20 \times 20}{20}\) = 20 metres.(u\(^2\)s cancel out)
Thus H\(_{max}\) attained = 20metres.
Ajụjụ 2 Ripọtì
(a)(i) What is a thermometric liquid?
(ii) State the reason for the following design features of a clinical thermometer. I. Narrow bore: II. Thin wall of the bulb
(b) Distinguish between heat and temperature of an object in terms of the energy of a particle
(c) Explain why evaporation leads to cooling
(d) A kettle rated 2000W, contains water at 20ºC. The kettle is switched on and after two minutes, the water starts boiling. After another six minutes, 45% of the water in the kettle boils away. (i) Determine the specific latent heat of the vaporization of the water (ii) State one assumption made in your calculation 9d(i) above
(a) A thermometric liquid is a liquid used in thermometers to measure temperature. It expands and contracts uniformly with changes in temperature, allowing for accurate readings.
(ii) I. Narrow Bore
Reason: Allows precise measurement by controlling the movement of the thermometric liquid, ensuring quick and accurate readings
II. Thin Wall of the Bulb
Reason: Enhances thermal conductivity, enabling faster heat transfer for quicker and more accurate temperature readings.
(b) Heat is a measure of the change in total internal energy in a body while temperature is a measure of the average kinetic energy of a molecule of the body.
(c) Evaporation leads to cooling for these reasons:
Energy Absorption: Molecules at the surface absorb energy to break free, often from the liquid and its surroundings.
Loss of High-Energy Molecules: Higher-energy molecules evaporate, reducing the average kinetic energy of the remaining liquid.
Temperature Decrease: As the average kinetic energy drops, the temperature of the liquid decreases, resulting in cooling.
(d)(i) Given: P = 2000W, \(\theta\) = 20ºC, t = 2 mins = 120 secs.
P x t = mc\(\Delta\)\(\theta\)
m = \(\frac{P \times t}{c \times \Delta \theta}\)
m = \(\frac{2000 \times 120}{4200 \times (100 - 20)}\) = 0.714 kg
Also, Pt = ml
l = \(\frac{Pt}{m}\)
where m = 45% of 0.714, t = 6mins = 360secs
l = \(\frac{2000 \times 60 \times 6}{0.714 \times 0.45}\) = 2.29 x 10\(^6\)Jkg\(^{-1}\)
(d)(i) Assumption made: there is no loss of heat to the surroundings, and the heat capacity of the material of the kettle is negligible.
Akọwa Nkọwa
(a) A thermometric liquid is a liquid used in thermometers to measure temperature. It expands and contracts uniformly with changes in temperature, allowing for accurate readings.
(ii) I. Narrow Bore
Reason: Allows precise measurement by controlling the movement of the thermometric liquid, ensuring quick and accurate readings
II. Thin Wall of the Bulb
Reason: Enhances thermal conductivity, enabling faster heat transfer for quicker and more accurate temperature readings.
(b) Heat is a measure of the change in total internal energy in a body while temperature is a measure of the average kinetic energy of a molecule of the body.
(c) Evaporation leads to cooling for these reasons:
Energy Absorption: Molecules at the surface absorb energy to break free, often from the liquid and its surroundings.
Loss of High-Energy Molecules: Higher-energy molecules evaporate, reducing the average kinetic energy of the remaining liquid.
Temperature Decrease: As the average kinetic energy drops, the temperature of the liquid decreases, resulting in cooling.
(d)(i) Given: P = 2000W, \(\theta\) = 20ºC, t = 2 mins = 120 secs.
P x t = mc\(\Delta\)\(\theta\)
m = \(\frac{P \times t}{c \times \Delta \theta}\)
m = \(\frac{2000 \times 120}{4200 \times (100 - 20)}\) = 0.714 kg
Also, Pt = ml
l = \(\frac{Pt}{m}\)
where m = 45% of 0.714, t = 6mins = 360secs
l = \(\frac{2000 \times 60 \times 6}{0.714 \times 0.45}\) = 2.29 x 10\(^6\)Jkg\(^{-1}\)
(d)(i) Assumption made: there is no loss of heat to the surroundings, and the heat capacity of the material of the kettle is negligible.
Ajụjụ 3 Ripọtì
PART2
(a) State the effect of increasing temperature on the viscosity of a: (i) liquid, (ii) gas
(b) State two factors that determine the magnitude of a moment of a force
(c) A uniform stick AB of length, L, and mass, m, is balanced horizontally on a knife edge 10.0cm from A when an object of 400 g is suspended at A. When the knife edge is moved 5 cm further, the object has to be moved to a point 9.00 cm from A for the stick to balance.
(i) Represent the balance system with a suitable diagram
(ii) Determine the: I. mass, m of the stick; II. length, L.
(d) Explain in terms of air molecules why pressure at the top of a high mountain is less than at sea level.
(e) Mercury of density 13.6 x 10\(^3\)kgm\(^{-3}\) is poured in a container of uniform cross-sectional area 40cm \(^2\) to a height of 20 cm. The total pressure exerted on the base of the container is 1.42 x 10\(^5\)P. Calculate the: (i) mass of the mercury in the container and (ii) pressure exerted at the surface of the mercury. [g = 10ms\(^{-2}\)
(ai) Increasing the temperature of a liquid decreases the viscosity of the liquid. (aii) Increasing the temperature of a gas increases the viscosity of the gas
(b) Factors that determine the magnitude of the moment of a force about a point are: 1. Angle between the distance from the turning point and the line of action of the force. 2. magnitude of the applied force. 3. perpendicular distance from the pivot to the line of action of the applied force.
(c) SEE THE DIAGRAM ABOVE.
(ii) Taking a moment about the support: clockwise moment = anti-clockwise moment
400 x 10 = m x (\(\frac{L}{2}\) - 10) -- ------- from figure 1 above
4000 = m x (\(\frac{L}{2}\) - 10) - - - -- - - (i)
400 x 6 = m x (\(\frac{L}{2}\) - 15) - - - - -- - - -(ii)
divide eqn i by ii
\(\frac{4000}{2400}\) = \(\frac{m \times \frac{L}{2} - 10)}{m \times \frac{L}{2} - 15)}\)
\(\frac{5}{3}\) = \(\frac{\frac{L}{2} - 10)}{\frac{L}{2} - 15)}\)
3 x (\(\frac{L}{2}\) - 10) = 5 x (\(\frac{L}{2}\) - 15)
\(\frac{3L}{2}\) - 30) = \(\frac{5L}{2}\) - 75)
\(\frac{5L}{2}\) - \(\frac{3L}{2}\) = 75 - 30 = 45
L = \(\frac{2L}{2}\) = 45 cm
From equation (ii) 400 x 6 = m x (\(\frac{L}{2}\) - 15)
2400 = m x (\(\frac{45}{2}\) - 15)
2400 = m x 7.5
m = \(\frac{2400}{7.5}\) = 320 g
Therefore, I. m = 320g and II. L = 45 cm
(d) Reasons for less pressure at the top of a mountain:
The lower pressure at the top of a mountain is due to the reduced number of air molecules in the atmosphere above that elevation, resulting in a lighter air column and fewer molecular collisions, which together create lower atmospheric pressure compared to sea level.
(e)(i) Density = \(\rho\) = \(\frac{m}{v}\)
m = v x \(\rho\), but v = A x h
m = A x h x \(\rho\) = 13.6 x 10\(^3\) x 40 x 20 x 10\(^{-6}\) = 10.88kg
(e)(ii) pressure exerted by the surface of the mercury
P\(_t\) = P\(_{atm}\) + pressure of mercury
P\(_{atm}\) = P\(_t\) - \(\rho\) x h x g = 1.42 x 10\(^5\) - 13.6 x 10\(^3\) x 0.2 x 10 = 1.148 x 10\(^5\)Pa
Akọwa Nkọwa
(ai) Increasing the temperature of a liquid decreases the viscosity of the liquid. (aii) Increasing the temperature of a gas increases the viscosity of the gas
(b) Factors that determine the magnitude of the moment of a force about a point are: 1. Angle between the distance from the turning point and the line of action of the force. 2. magnitude of the applied force. 3. perpendicular distance from the pivot to the line of action of the applied force.
(c) SEE THE DIAGRAM ABOVE.
(ii) Taking a moment about the support: clockwise moment = anti-clockwise moment
400 x 10 = m x (\(\frac{L}{2}\) - 10) -- ------- from figure 1 above
4000 = m x (\(\frac{L}{2}\) - 10) - - - -- - - (i)
400 x 6 = m x (\(\frac{L}{2}\) - 15) - - - - -- - - -(ii)
divide eqn i by ii
\(\frac{4000}{2400}\) = \(\frac{m \times \frac{L}{2} - 10)}{m \times \frac{L}{2} - 15)}\)
\(\frac{5}{3}\) = \(\frac{\frac{L}{2} - 10)}{\frac{L}{2} - 15)}\)
3 x (\(\frac{L}{2}\) - 10) = 5 x (\(\frac{L}{2}\) - 15)
\(\frac{3L}{2}\) - 30) = \(\frac{5L}{2}\) - 75)
\(\frac{5L}{2}\) - \(\frac{3L}{2}\) = 75 - 30 = 45
L = \(\frac{2L}{2}\) = 45 cm
From equation (ii) 400 x 6 = m x (\(\frac{L}{2}\) - 15)
2400 = m x (\(\frac{45}{2}\) - 15)
2400 = m x 7.5
m = \(\frac{2400}{7.5}\) = 320 g
Therefore, I. m = 320g and II. L = 45 cm
(d) Reasons for less pressure at the top of a mountain:
The lower pressure at the top of a mountain is due to the reduced number of air molecules in the atmosphere above that elevation, resulting in a lighter air column and fewer molecular collisions, which together create lower atmospheric pressure compared to sea level.
(e)(i) Density = \(\rho\) = \(\frac{m}{v}\)
m = v x \(\rho\), but v = A x h
m = A x h x \(\rho\) = 13.6 x 10\(^3\) x 40 x 20 x 10\(^{-6}\) = 10.88kg
(e)(ii) pressure exerted by the surface of the mercury
P\(_t\) = P\(_{atm}\) + pressure of mercury
P\(_{atm}\) = P\(_t\) - \(\rho\) x h x g = 1.42 x 10\(^5\) - 13.6 x 10\(^3\) x 0.2 x 10 = 1.148 x 10\(^5\)Pa
Ajụjụ 4 Ripọtì
State three uses of ferromagnetic materials
The uses of ferromagnetic materials are:
(i) Electromagnets
(ii) Magnetic Storage
(iii) Magnetic Sensors
(iv) Magnetic Shielding
(v) Permanent Magnets
(vi) Inductive Components, others are transformers, electric bells, electric generators, etc.
Akọwa Nkọwa
The uses of ferromagnetic materials are:
(i) Electromagnets
(ii) Magnetic Storage
(iii) Magnetic Sensors
(iv) Magnetic Shielding
(v) Permanent Magnets
(vi) Inductive Components, others are transformers, electric bells, electric generators, etc.
Ajụjụ 5 Ripọtì
An electron of mass, m, and charge, e moves through the electric field of potential difference V\(_o\) with a speed, v. Show that de Broglie wavelength associated with the electron is given as \(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\), where h is the Plank's constant.
To show that \(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\)
The de broglie wavelength \(\lambda\), is given by
\(\lambda\) = \(\frac{\text{h}}{\text{p}}\)
where p = mv
\(\lambda\) = \(\frac{\text{h}}{\text{mv}}\)
When an electron is accelerated through a potential difference V\(_0\), it gains kinetic energy given K.E.= eV\(_0\)
The kinetic energy can also be expressed in terms of the mass m of the electron and its speed v: K.E = \(\frac{1}{2}\)mv\(^2\)
K.E = eV\(_0\) = \(\frac{1}{2}\)mv\(^2\)
v = \(\sqrt{\frac{2eV_o}{m}}\)
Substituting v = \(\sqrt{\frac{2eV_o}{m}}\) into the de Broglie equation
\(\lambda\) = \(\frac{\text{h}}{m \times \sqrt{\frac{2eV_o}{m}}}\) = \(\frac{\text{h}}{\sqrt{\frac{2eV_o m^2}{m}}}\)
\(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\)
Note: use the fact that \(\sqrt{m}\) x \(\sqrt{m}\) = m
Akọwa Nkọwa
To show that \(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\)
The de broglie wavelength \(\lambda\), is given by
\(\lambda\) = \(\frac{\text{h}}{\text{p}}\)
where p = mv
\(\lambda\) = \(\frac{\text{h}}{\text{mv}}\)
When an electron is accelerated through a potential difference V\(_0\), it gains kinetic energy given K.E.= eV\(_0\)
The kinetic energy can also be expressed in terms of the mass m of the electron and its speed v: K.E = \(\frac{1}{2}\)mv\(^2\)
K.E = eV\(_0\) = \(\frac{1}{2}\)mv\(^2\)
v = \(\sqrt{\frac{2eV_o}{m}}\)
Substituting v = \(\sqrt{\frac{2eV_o}{m}}\) into the de Broglie equation
\(\lambda\) = \(\frac{\text{h}}{m \times \sqrt{\frac{2eV_o}{m}}}\) = \(\frac{\text{h}}{\sqrt{\frac{2eV_o m^2}{m}}}\)
\(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\)
Note: use the fact that \(\sqrt{m}\) x \(\sqrt{m}\) = m
Ajụjụ 6 Ripọtì
(a) State two uses of a polar satellite.
(b) What does the slope of a graph of tensile stress against tensile strain represent?
(a) The uses of polar satellites include:
(i) for location identification
(ii) to collect data on climate change
(iii) observe and predict a natural disaster
(iv) use for communication.
(b) The slope of a graph of tensile stress against tensile strain represents the modulus of elasticity, also known as Young's modulus. Young's modulus quantifies the stiffness of a material. A steeper slope indicates a stiffer material, while a flatter slope indicates a more flexible material.
Akọwa Nkọwa
(a) The uses of polar satellites include:
(i) for location identification
(ii) to collect data on climate change
(iii) observe and predict a natural disaster
(iv) use for communication.
(b) The slope of a graph of tensile stress against tensile strain represents the modulus of elasticity, also known as Young's modulus. Young's modulus quantifies the stiffness of a material. A steeper slope indicates a stiffer material, while a flatter slope indicates a more flexible material.
Ajụjụ 7 Ripọtì
The fractional change in length produced in an elastic material of spring constant 680Nm\(^{-1}\) when a force of 306N is applied to stretch it is 1.5. Calculate the original length of the material.
From Hooke's law, F = kx where x = extension, k = force constant
K = 680Nm\(^{-1}\), F = 306N
x = \(\frac{\text{F}}{\text{x}}\) = \(\frac{306}{680}\) ≈ 0.450m
Fractional Change in Length: Given the fractional change is 1.5:
1.5 = \(\frac{\text{x}}{\text{L}}\)
L = \(\frac{0.450}{1.5}\) = 0.30m.
Akọwa Nkọwa
From Hooke's law, F = kx where x = extension, k = force constant
K = 680Nm\(^{-1}\), F = 306N
x = \(\frac{\text{F}}{\text{x}}\) = \(\frac{306}{680}\) ≈ 0.450m
Fractional Change in Length: Given the fractional change is 1.5:
1.5 = \(\frac{\text{x}}{\text{L}}\)
L = \(\frac{0.450}{1.5}\) = 0.30m.
Ajụjụ 8 Ripọtì
(a) Derive the dimension of surface tension.
(b) Name the instrument used to measure the force of gravity at a place
(a) SURFACE TENSION = \(\frac{\text{force}}{\text{length}}\)
FORCE = mass x accel. due to gravity
Dimension of force = M.LT\(^{-2}\)
Dimension of length = L
Dimension of surface area = \(\frac{MLT^{-2}}{L}\) = MT\(^{-2}\)
(b) The instruments used for measuring the force of gravity at a place are:
(1) Gravimeter, (2) Spring balance, (3) Gravitational accelerometer.
Akọwa Nkọwa
(a) SURFACE TENSION = \(\frac{\text{force}}{\text{length}}\)
FORCE = mass x accel. due to gravity
Dimension of force = M.LT\(^{-2}\)
Dimension of length = L
Dimension of surface area = \(\frac{MLT^{-2}}{L}\) = MT\(^{-2}\)
(b) The instruments used for measuring the force of gravity at a place are:
(1) Gravimeter, (2) Spring balance, (3) Gravitational accelerometer.
Ị ga-achọ ịga n'ihu na omume a?