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Question 1 Report
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.
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.
Question 2 Report
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.
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.
Question 3 Report
(a) Mention three facts about photoelectric effect
(b) An electric magnetic radiation source of power 6 x 10\(^{-3}\) W emits 1 x 10\(^{16}\) photons per second. The most energetic photo electron ejected from a metal surface is stopped by a potential difference of 2.2V. Calculate the work function of the metal. [ mass of (photon) electrons 9.1 x 10\(^{-31}\) kg, e = 1.6 x 10\(^{-19}\) C]
(c) State two factors on which the activity of a radioactive sample depends
(d) Cobalt-60 source has an activity of 2.0 x 10\(^6\) Bq and a half-life of 1.8 x 10\(^8\)s. Calculate the number of radioisotope nuclei in the source.
(a) electrons are only emitted when the photon frequency exceeds the threshold frequency,
- the energy of each photon is proportional to its frequency,
- the emission of photoelectrons depends on the frequency of incident light on the metal,
- the intensity of light does not affect the energy of the electrons emitted/photoelectrons,
- the energy of the photoelectron depends on the frequency of incident radiation.
(b) Given: P = 6 x 10\(^{-3}\) W, no of photons per seconds = 1 x 10\(^{16}\), m\(_e\) = 9.1 x 10\(^{-31}\) kg, e = 1.6 x 10\(^{-19}\) C, W\(_o\) = ?
Photon Energy E = \(\frac{\text{power}}{\text{no of photons per second}}\) = \(\frac{\text{P}}{\text{n}}\)
= \(\frac{6.0 \times 10^{-3}}{1 \times 10^{16}}\)
= 6 x 10\(^{-19}\)J
But K.E = E + W\(_o\)
W\(_o\) = E - K.E = E - eV
W\(_o\) = 6 x 10\(^{-19}\) - 2.2 x (1.6 x 10\(^{-19}\)) = 2.48 x 10\(^{-9}\)J
(c) Factors on which radioactive activity depends are: (i) decay constant, (ii) number of unstable nuclei present at a given time (iii) time of the decay process (iv) environmental conditions, (v) type of radioisotope, etc
(d) Given: A = 2.0 x 10\(^6\)Bq, t\(_{\frac{1}{2}}\) = 1.8 x 10\(^8\)s
A = \(\lambda\)N, where \(\lambda\) = decay constant, N = number of radioactive nuclei, A = activity.
But \(\lambda\) = \(\frac{In(2)}{t_{\frac{1}{2}}}\)
\(\lambda\) = \(\frac{In(2)}{1.8 \times 10^8}\) ≈ \(\frac{0.693}{1.8 \times 10^8}\) ≈ 3.85 x 10\(^{-9}\)s\(^{-1}\)
Recall, A = \(\lambda\)N
N = \(\frac{A}{\lambda}\) = \(\frac{2.0 \times 10^6}{3.85 \times 10^{-9}}\) = 5.2 x 10\(^{14}\) nuclei
(a) electrons are only emitted when the photon frequency exceeds the threshold frequency,
- the energy of each photon is proportional to its frequency,
- the emission of photoelectrons depends on the frequency of incident light on the metal,
- the intensity of light does not affect the energy of the electrons emitted/photoelectrons,
- the energy of the photoelectron depends on the frequency of incident radiation.
(b) Given: P = 6 x 10\(^{-3}\) W, no of photons per seconds = 1 x 10\(^{16}\), m\(_e\) = 9.1 x 10\(^{-31}\) kg, e = 1.6 x 10\(^{-19}\) C, W\(_o\) = ?
Photon Energy E = \(\frac{\text{power}}{\text{no of photons per second}}\) = \(\frac{\text{P}}{\text{n}}\)
= \(\frac{6.0 \times 10^{-3}}{1 \times 10^{16}}\)
= 6 x 10\(^{-19}\)J
But K.E = E + W\(_o\)
W\(_o\) = E - K.E = E - eV
W\(_o\) = 6 x 10\(^{-19}\) - 2.2 x (1.6 x 10\(^{-19}\)) = 2.48 x 10\(^{-9}\)J
(c) Factors on which radioactive activity depends are: (i) decay constant, (ii) number of unstable nuclei present at a given time (iii) time of the decay process (iv) environmental conditions, (v) type of radioisotope, etc
(d) Given: A = 2.0 x 10\(^6\)Bq, t\(_{\frac{1}{2}}\) = 1.8 x 10\(^8\)s
A = \(\lambda\)N, where \(\lambda\) = decay constant, N = number of radioactive nuclei, A = activity.
But \(\lambda\) = \(\frac{In(2)}{t_{\frac{1}{2}}}\)
\(\lambda\) = \(\frac{In(2)}{1.8 \times 10^8}\) ≈ \(\frac{0.693}{1.8 \times 10^8}\) ≈ 3.85 x 10\(^{-9}\)s\(^{-1}\)
Recall, A = \(\lambda\)N
N = \(\frac{A}{\lambda}\) = \(\frac{2.0 \times 10^6}{3.85 \times 10^{-9}}\) = 5.2 x 10\(^{14}\) nuclei
Question 4 Report
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
(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
Question 5 Report
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
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
Question 6 Report
(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.
(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.
Question 7 Report
(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.
(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.
Question 8 Report
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.
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.
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