WAEC SSCE - Physics - 2023 (Essay)
Pregunta 1 Informe
ai. Why are parabolic mirrors suitable for use in the headlamps of vehicles?
ii. Draw a ray diagram to illustrate the answer in 10(a)(i).
bi. State two applications of echoes.
ii. An observer standing at a point, P, on the same horizontal ground as the foot, H, of a tower, shouts, and 1.20 s later, he hears the echo. He then moved to another point, Q, 40 m from P, and shouted again but the echo was heard after 1.45 s. Calculate the:
I. distance between P and H;
II. speed of sound in air.
ci. Define the term absolute refractive index of a medium.
ii. A piece of coin falls accidentally into a tank containing two immiscible liquids A and B as illustrated in Fig. 10.0 above.
Calculate the displacement of the coin when viewed vertically from above.
[refractive index of A = 1.3, refractive index of B = 1.4]
ai. Parabolic mirrors, due to their ability to efficiently collect and focus incoming parallel light rays, are suitable for use in vehicle headlamps, ensuring a concentrated and directed beam of light for improved visibility and reduced glare.
ii. See diagram above.
bi. – Sonar Systems: Sonar (Sound Navigation and Ranging) systems use echoes to determine the depth and location of objects underwater. A sound signal is emitted, and the time it takes for the signal to bounce off an object and return as an echo is used to calculate the distance to the object. Sonar is commonly used in marine navigation, fishing, and underwater mapping.
- Medical Imaging: In medical ultrasound imaging, echoes are used to create images of the interior of the body. High-frequency sound waves are directed into the body, and the echoes that bounce back from different tissues are used to generate detailed images of organs, blood vessels, and developing fetuses during pregnancy.
- Architectural Design: Architects and engineers use echoes to evaluate the acoustics of buildings and design spaces with optimal sound qualities. The controlled use of echoes can enhance the sound in concert halls, auditoriums, and theaters, providing a more enjoyable listening experience for audiences.
- Geological Surveys: Seismic surveys use echoes generated by controlled explosions or mechanical sources to map subsurface geological features. By analyzing the time it takes for seismic waves to reflect off different rock layers, geologists can infer the composition and structure of the Earth's subsurface.
- Surveying and Distance Measurement: Echoes are used in surveying and distance measurement equipment like total stations and laser rangefinders. A laser or light signal is emitted, and the time it takes for the light to bounce off a target and return is used to measure distances accurately.
- Echo Sounding in Navigation: In addition to sonar, echo sounding is used in navigation to determine water depth, especially in shallow or coastal areas. A sound signal is sent to the seabed, and the time it takes for the echo to return helps sailors and navigators avoid underwater hazards.
- Musical Effects: Musicians and audio engineers often use echoes and reverb effects in music production. These effects add depth and dimension to music by simulating the reflection of sound in different environments. Echoes and reverb are commonly used in recording studios, live concerts, and digital audio effects.
See diagram above.
ii. V = \(\frac{2d}{t}\)
Case1: V = \(\frac{2d}{1.2}\)
Case2: V = \(\frac{2(40 + d)}{1.45}\)
Equating case 1 and 2
\(\frac{2d}{1.2}\) = \(\frac{2(40 + d)}{1.45}\)
2.9d = 96 + 2.4d
2.9d - 2.4d = 96
0.5d = 96
d = \(\frac{96}{0.5}\)
= 192m
∴ Distance between P and H = 192m
put d = 192m into case(1)
V = \(\frac{2d}{1.2}\)
V = \(\frac{2\times192}{1.2}\) = 320m/s
ci. The absolute refractive index of a medium is the ratio of the speed of light in a vacuum to the speed of light in the medium. It is a measure of how much a medium slows down light.
ii. Displacement = real depth - \(\frac{real depth}{refractive index }\) ( n = \(\frac{real depth}{apparentdepth }\))
= real depth( 1 - \(\frac{1}{n}\))
For liquid A, displacement,d = 8 (1 - \(\frac{1}{1.3}\)) = 9.23cm
For liquid B, displacement,d = 40( 1 - \(\frac{1}{1.4}\)) = 2.29cm
∴ Total displacement = 9.23 + 2.29 = 11.52cm
∴ the apparent position of the coin is 11.52cm from the bottom.
Detalles de la respuesta
ai. Parabolic mirrors, due to their ability to efficiently collect and focus incoming parallel light rays, are suitable for use in vehicle headlamps, ensuring a concentrated and directed beam of light for improved visibility and reduced glare.
ii. See diagram above.
bi. – Sonar Systems: Sonar (Sound Navigation and Ranging) systems use echoes to determine the depth and location of objects underwater. A sound signal is emitted, and the time it takes for the signal to bounce off an object and return as an echo is used to calculate the distance to the object. Sonar is commonly used in marine navigation, fishing, and underwater mapping.
- Medical Imaging: In medical ultrasound imaging, echoes are used to create images of the interior of the body. High-frequency sound waves are directed into the body, and the echoes that bounce back from different tissues are used to generate detailed images of organs, blood vessels, and developing fetuses during pregnancy.
- Architectural Design: Architects and engineers use echoes to evaluate the acoustics of buildings and design spaces with optimal sound qualities. The controlled use of echoes can enhance the sound in concert halls, auditoriums, and theaters, providing a more enjoyable listening experience for audiences.
- Geological Surveys: Seismic surveys use echoes generated by controlled explosions or mechanical sources to map subsurface geological features. By analyzing the time it takes for seismic waves to reflect off different rock layers, geologists can infer the composition and structure of the Earth's subsurface.
- Surveying and Distance Measurement: Echoes are used in surveying and distance measurement equipment like total stations and laser rangefinders. A laser or light signal is emitted, and the time it takes for the light to bounce off a target and return is used to measure distances accurately.
- Echo Sounding in Navigation: In addition to sonar, echo sounding is used in navigation to determine water depth, especially in shallow or coastal areas. A sound signal is sent to the seabed, and the time it takes for the echo to return helps sailors and navigators avoid underwater hazards.
- Musical Effects: Musicians and audio engineers often use echoes and reverb effects in music production. These effects add depth and dimension to music by simulating the reflection of sound in different environments. Echoes and reverb are commonly used in recording studios, live concerts, and digital audio effects.
See diagram above.
ii. V = \(\frac{2d}{t}\)
Case1: V = \(\frac{2d}{1.2}\)
Case2: V = \(\frac{2(40 + d)}{1.45}\)
Equating case 1 and 2
\(\frac{2d}{1.2}\) = \(\frac{2(40 + d)}{1.45}\)
2.9d = 96 + 2.4d
2.9d - 2.4d = 96
0.5d = 96
d = \(\frac{96}{0.5}\)
= 192m
∴ Distance between P and H = 192m
put d = 192m into case(1)
V = \(\frac{2d}{1.2}\)
V = \(\frac{2\times192}{1.2}\) = 320m/s
ci. The absolute refractive index of a medium is the ratio of the speed of light in a vacuum to the speed of light in the medium. It is a measure of how much a medium slows down light.
ii. Displacement = real depth - \(\frac{real depth}{refractive index }\) ( n = \(\frac{real depth}{apparentdepth }\))
= real depth( 1 - \(\frac{1}{n}\))
For liquid A, displacement,d = 8 (1 - \(\frac{1}{1.3}\)) = 9.23cm
For liquid B, displacement,d = 40( 1 - \(\frac{1}{1.4}\)) = 2.29cm
∴ Total displacement = 9.23 + 2.29 = 11.52cm
∴ the apparent position of the coin is 11.52cm from the bottom.
Pregunta 2 Informe
a. What is fibre optics?
b. State two reasons why optical fibres are preferred to copper cables in the telecommunication industry.
a. Fiber optics is a technology that uses light to transmit data over long distances. Fiber optic cables are made of thin, flexible strands of glass or plastic that can carry a large amount of data at very high speeds.
b. - Higher bandwidth: Optical fibers can carry a much larger amount of data than copper cables. This is because light travels much faster than electricity and optical fibers can carry more light signals than copper cables can carry electrical signals.
- Longer distances: Optical fibers can transmit data over longer distances than copper cables without signal loss. This is because light signals are less attenuated in optical fibers than electrical signals are in copper cables.
- Immunity to interference: Optical fibers are immune to electrical interference, which can cause problems with copper cables. This is because optical fibers transmit light signals, which are not affected by electrical fields.
- Security: Optical fibers are more secure than copper cables because they are difficult to tap. This is because light signals cannot be detected without damaging the optical fiber.
- Durability: Optical fibers are more durable than copper cables. They are less susceptible to damage from moisture, temperature fluctuations, and physical impact.
- Cost-effectiveness: Optical fibers are becoming more cost-effective than copper cables, especially for high-bandwidth applications. This is because the cost of manufacturing optical fibers has decreased significantly in recent years.
Detalles de la respuesta
a. Fiber optics is a technology that uses light to transmit data over long distances. Fiber optic cables are made of thin, flexible strands of glass or plastic that can carry a large amount of data at very high speeds.
b. - Higher bandwidth: Optical fibers can carry a much larger amount of data than copper cables. This is because light travels much faster than electricity and optical fibers can carry more light signals than copper cables can carry electrical signals.
- Longer distances: Optical fibers can transmit data over longer distances than copper cables without signal loss. This is because light signals are less attenuated in optical fibers than electrical signals are in copper cables.
- Immunity to interference: Optical fibers are immune to electrical interference, which can cause problems with copper cables. This is because optical fibers transmit light signals, which are not affected by electrical fields.
- Security: Optical fibers are more secure than copper cables because they are difficult to tap. This is because light signals cannot be detected without damaging the optical fiber.
- Durability: Optical fibers are more durable than copper cables. They are less susceptible to damage from moisture, temperature fluctuations, and physical impact.
- Cost-effectiveness: Optical fibers are becoming more cost-effective than copper cables, especially for high-bandwidth applications. This is because the cost of manufacturing optical fibers has decreased significantly in recent years.
Pregunta 3 Informe
a. State the function of each of the following parts of a modern x-ray tube: i. heater; ii. high tension source; iii. cooling fins.
b. State one reason for each of the following design features of a modern x-ray tube: i. the glass envelope is highly evacuated; ii. the target is a metal of very high melting point; iii. the cooling fins are located outside the glass envelope.
c. In a nuclear fission reaction, a nuclide \(^{235}U_{92}\) is bombarded with a neutron to produce \(^{93}Kr_{36}\) and \(^{141}Ba_{56}\) with additional neutrons, the energy involved in the process is Q.
[mass of \(^{235}U_{92}\) = 235.044 u, mass of \(^{93}Kr_ {36}\) = 91.898 u, mass of \(^{141}Ba_ {56}\) = 140.914 u,
mass of neutron = 1.009u, 1u = \(1.66 \times 10^{-27}\) kg, c = 3.0 × \(10^8ms^1\)]
i. Write down the balanced nuclear reaction equation for the process.
ii. State with reason whether Q is absorbed or released in the process.
iii. Calculate the value of Q in joules.
ai. The heater, often referred to as the filament, is responsible for emitting electrons when heated.
ii. The high tension source, or high voltage generator, provides the high voltage necessary to accelerate the electrons emitted from the filament towards the anode (target) of the X-ray tube.
iii. Cooling fins are designed to dissipate the heat generated during the operation of the X-ray tube.
bi. The X-ray tube is highly evacuated so that the accelerated electrons can get to their target without losing much of their energy.
ii. The target is a metal with a very high melting point to withstand the high temperature generated when the electrons strike the target.
iii. The cooling fins are placed outside the glass envelope to ensure that the heat generated within the X-ray tube is efficiently transferred to the external environment.
ci. \(^{235}U_{92} + ^1_0n → ^{93}Kr_{36} + ^{141}Ba_{56} + 2^1_0n + Q\)
ii. Energy, Q, is released because the total mass on the left-hand side (reactant) is more than the total mass on the right-hand side (product). So, the energy is released in the form of a reduction in total mass. This missing mass is known as the 'mass defect' and it accounts for the energy released. Also, it's induced fission and nuclear fission is a process in which an unstable nucleus splits into two other lighter nuclei together with several neutrons and is accompanied by the release of energy.
iii. E = ∆\(mc^2\) where ∆m is the mass defect.
Total mass on the LHS:
235.044 u + 1.009 u = 236.053 u
Total mass on the RHS:
91.898 u + 140.914 u + 2(1.009 u) = 234.83 u
Mass defect, ∆m = 236.053 u - 234.83 u = 1.223 u
∆m = 1.223 × 1.66 × \(10^{-27}\) = 2.03 × \(10^{-27}\) kg
So,
E = 2.03 × \(10^{-27}\) × (3 × \(10^8)^2\)
= 1.827 × \(10^{-10}\) J
Detalles de la respuesta
ai. The heater, often referred to as the filament, is responsible for emitting electrons when heated.
ii. The high tension source, or high voltage generator, provides the high voltage necessary to accelerate the electrons emitted from the filament towards the anode (target) of the X-ray tube.
iii. Cooling fins are designed to dissipate the heat generated during the operation of the X-ray tube.
bi. The X-ray tube is highly evacuated so that the accelerated electrons can get to their target without losing much of their energy.
ii. The target is a metal with a very high melting point to withstand the high temperature generated when the electrons strike the target.
iii. The cooling fins are placed outside the glass envelope to ensure that the heat generated within the X-ray tube is efficiently transferred to the external environment.
ci. \(^{235}U_{92} + ^1_0n → ^{93}Kr_{36} + ^{141}Ba_{56} + 2^1_0n + Q\)
ii. Energy, Q, is released because the total mass on the left-hand side (reactant) is more than the total mass on the right-hand side (product). So, the energy is released in the form of a reduction in total mass. This missing mass is known as the 'mass defect' and it accounts for the energy released. Also, it's induced fission and nuclear fission is a process in which an unstable nucleus splits into two other lighter nuclei together with several neutrons and is accompanied by the release of energy.
iii. E = ∆\(mc^2\) where ∆m is the mass defect.
Total mass on the LHS:
235.044 u + 1.009 u = 236.053 u
Total mass on the RHS:
91.898 u + 140.914 u + 2(1.009 u) = 234.83 u
Mass defect, ∆m = 236.053 u - 234.83 u = 1.223 u
∆m = 1.223 × 1.66 × \(10^{-27}\) = 2.03 × \(10^{-27}\) kg
So,
E = 2.03 × \(10^{-27}\) × (3 × \(10^8)^2\)
= 1.827 × \(10^{-10}\) J
Pregunta 4 Informe
ai. State the reason why simple harmonic motion is periodic.
ii. State two factors that affect the period of oscillation of a simple pendulum.
iii. Sketch a graph of the total mechanical energy, E, against displacement, y, for the motion of a simple pendulum from one extreme position to the other.
b. The diagram above illustrates an oscillatory pendulum. Calculate the work done in raising the pendulum to point B, if the mass of the bob is 50 g.
[g = \(10 ms^2\)] see the figure above
c. A spiral spring of spring constant, k, and natural length, l, has a scale pan of mass 0.04 kg hanging on its lower end while the upper end is firmly fixed to a support. When an object of mass 0.20 kg is placed on the scale pan, the length of the spring becomes 0.055 m and when the object is replaced with another object of mass 0.28 kg, the length of the spring becomes 0.065 m. Calculate the values of k and l.
[g = \(10 ms^2\)]
ai. Simple harmonic motion is periodic because of the restoring force which is proportional to the displacement, acts in the opposite direction to the motion
ii. ♦ Length of the pendulum ♦ Strength of the gravitational field
iii. See the figure above.
b. m = 50g = 0.05kg, g = 10 \(ms^2\), h = 10cm = 0.1m, P.E = ?
P.E = mgh = 0.05 × 10 × 0.1
∴ P.E = 0.05 J
c. Let the extension be ∆l
F = k ∆l = k(\(l_n\) - l )
where k is the spring constant, ln is the new length, l is the natural length, and F = mg.
Total mass = mass of pan + added mass
In the first case,
\(m_1\) = 0.04 + 0.20 = 0.24 kg
\(F_1\) = 0.24 × 10 = 2.4 N
∆l = 0.055 - l
Since F = k ∆l,
2.4 = k(0.055 - l )
2.4 = 0.055k - l k --- (i)
In the second case,
\(m_2\) = 0.04 + 0.28 = 0.32 kg
\(F_2\) = 0.32 × 10 = 3.2 N
∆l = 0.065 - l
Since F = k ∆l,
3.2 = k(0.065 - l )
3.2 = 0.065k - l k --- (ii)
Subtracting equation (i) from (ii)
⇒ 0.8 = 0.01k
k = \(\frac{0.8}{ 0.01}\)
Substitute (80) for k in equation (i)
⇒ 2.4 = 0.055(80) - 80 l
⇒ 2.4 = 4.4 - 80 l
⇒ 80 l = 4.4 - 2.4
⇒ 80 l = 2
l = \(\frac{2}{80}\)
Therefore, the value of k = 80 N/m and l = 0.025 m
Detalles de la respuesta
ai. Simple harmonic motion is periodic because of the restoring force which is proportional to the displacement, acts in the opposite direction to the motion
ii. ♦ Length of the pendulum ♦ Strength of the gravitational field
iii. See the figure above.
b. m = 50g = 0.05kg, g = 10 \(ms^2\), h = 10cm = 0.1m, P.E = ?
P.E = mgh = 0.05 × 10 × 0.1
∴ P.E = 0.05 J
c. Let the extension be ∆l
F = k ∆l = k(\(l_n\) - l )
where k is the spring constant, ln is the new length, l is the natural length, and F = mg.
Total mass = mass of pan + added mass
In the first case,
\(m_1\) = 0.04 + 0.20 = 0.24 kg
\(F_1\) = 0.24 × 10 = 2.4 N
∆l = 0.055 - l
Since F = k ∆l,
2.4 = k(0.055 - l )
2.4 = 0.055k - l k --- (i)
In the second case,
\(m_2\) = 0.04 + 0.28 = 0.32 kg
\(F_2\) = 0.32 × 10 = 3.2 N
∆l = 0.065 - l
Since F = k ∆l,
3.2 = k(0.065 - l )
3.2 = 0.065k - l k --- (ii)
Subtracting equation (i) from (ii)
⇒ 0.8 = 0.01k
k = \(\frac{0.8}{ 0.01}\)
Substitute (80) for k in equation (i)
⇒ 2.4 = 0.055(80) - 80 l
⇒ 2.4 = 4.4 - 80 l
⇒ 80 l = 4.4 - 2.4
⇒ 80 l = 2
l = \(\frac{2}{80}\)
Therefore, the value of k = 80 N/m and l = 0.025 m
Pregunta 5 Informe
a. Using the kinetic theory, explain the term diffusion of fluid molecules.
b. Name one phenomenon that demonstrates that light behaves as a:
i. wave ii. particles
a. According to the kinetic theory, diffusion in fluids occurs because of the random and continuous motion of molecules. This motion results in a net movement of the molecules from regions of higher concentration to regions to regions of lower concentration ultimately leading to the mixing of substances in the fluid i.e. the process continues until the concentration becomes uniform throughout the fluid. It's driven by the kinetic energy of the molecules and leads to the mixing of substances in fluids.
b.
i. Interference of light waves
ii .– Photo-electric effect
–Compton Effect
Detalles de la respuesta
a. According to the kinetic theory, diffusion in fluids occurs because of the random and continuous motion of molecules. This motion results in a net movement of the molecules from regions of higher concentration to regions to regions of lower concentration ultimately leading to the mixing of substances in the fluid i.e. the process continues until the concentration becomes uniform throughout the fluid. It's driven by the kinetic energy of the molecules and leads to the mixing of substances in fluids.
b.
i. Interference of light waves
ii .– Photo-electric effect
–Compton Effect
Pregunta 6 Informe
ai. Define the electric potential at a point in an electric field.
ii. An uncharged body, A, was charged electrostatically by a test charge, B, using the method of induction and the method of contact. State two differences between the two methods.
b. An important precaution during an electricity experiment is to open the circuit when no readings are being taken. Give two reasons for the stated precaution.
ci. Fig. 11.0 is a circuit diagram in which a coil of inductance, L, and a resistor of resistance, R, are connected to a variable alternating source of frequency, f.
The table shows the square of the impedance, \(Z^2\); corresponding to each value of \(ƒ^2\).
| \(ƒ^2\)/ \(Hz^2\) |
198.80 | 400.00 | 600.30 | 800.90 | 900.00 |
| \(Z^2\)/ \(Ω^2\) |
249.60 | 400.00 | 550.30 | 702.30 | 800.90 |
Write down the equation for Z in terms of \(f^2\), \(R^2\), and \(L^2\).
ii. Plot a graph \(Z^2 against \(f^2\) of and use it to determine the values of:
i. L
ii. R
[\(π^2\) = 10]
ai. Electric potential at a point is defined as the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field.
ii. Process of Charging:
Induction: Involves redistributing charges within an uncharged body without direct contact with a charged object.
Contact: Requires direct physical contact with a charged object to transfer charges.
- Nature of Final Charge:
Induction: Results in a charge of opposite polarity to the charged object initially brought close.
Contact: Results in a charge of the same polarity as the charged object with which contact is made.
b. - Safety: Opening the circuit prevents current from flowing, which reduces the risk of electric shock. This is especially important when working with high voltages or currents.
- Conservation of battery power: If the circuit is left closed, the current will continue to flow even when no readings are being taken. This can drain the battery and shorten its lifespan.
- Accurate readings: If the circuit is left closed, the current flowing through the circuit can cause the components to heat up. This can affect the accuracy of the readings.
- Protection of equipment: If the circuit is left closed, a surge of current could flow through the circuit if there is a fault. This could damage the equipment or even cause a fire.
c. \(Z^2 = R^2 + X{_L}^2\)
\(Z^2 = R^2 + (2πf_L)^2
\(Z^2 = R^2 + 4π^2f^2L^2\)
\(Z^2 = 4π^2f^2L^2 + R^2\)
∴ \(Z = \sqrt{4π^2L^2f^2 + R^2}\)
ii. from the equation
\(Z^2 = 4π^2f^2L^2 + R^2\)
using y = mx + c = y = \(Z^2\), x = \(f^2\)
m = \(4π^2L^2\) = slope of the grape
m = \(\frac{ Z_2^2 - Z_1^2}{f_2^2 - f_1^2} = \frac{ 700 - 200 }{ 800 - 140 }\)
\(4π^2L^2 = \frac{ 500}{ 660}\)
\(L^2 = \frac{ 500}{ 660} \times \frac{1}{ 4π^2} =\frac{ 500}{ 660} \times \frac{1}{4 \times 10}\) ( since \(π^2\) = 10)
\(L^2 = 0.0189\)
∴ L = \(\sqrt{ 0.0189}\) = 0.138H = 138mH
Detalles de la respuesta
ai. Electric potential at a point is defined as the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field.
ii. Process of Charging:
Induction: Involves redistributing charges within an uncharged body without direct contact with a charged object.
Contact: Requires direct physical contact with a charged object to transfer charges.
- Nature of Final Charge:
Induction: Results in a charge of opposite polarity to the charged object initially brought close.
Contact: Results in a charge of the same polarity as the charged object with which contact is made.
b. - Safety: Opening the circuit prevents current from flowing, which reduces the risk of electric shock. This is especially important when working with high voltages or currents.
- Conservation of battery power: If the circuit is left closed, the current will continue to flow even when no readings are being taken. This can drain the battery and shorten its lifespan.
- Accurate readings: If the circuit is left closed, the current flowing through the circuit can cause the components to heat up. This can affect the accuracy of the readings.
- Protection of equipment: If the circuit is left closed, a surge of current could flow through the circuit if there is a fault. This could damage the equipment or even cause a fire.
c. \(Z^2 = R^2 + X{_L}^2\)
\(Z^2 = R^2 + (2πf_L)^2
\(Z^2 = R^2 + 4π^2f^2L^2\)
\(Z^2 = 4π^2f^2L^2 + R^2\)
∴ \(Z = \sqrt{4π^2L^2f^2 + R^2}\)
ii. from the equation
\(Z^2 = 4π^2f^2L^2 + R^2\)
using y = mx + c = y = \(Z^2\), x = \(f^2\)
m = \(4π^2L^2\) = slope of the grape
m = \(\frac{ Z_2^2 - Z_1^2}{f_2^2 - f_1^2} = \frac{ 700 - 200 }{ 800 - 140 }\)
\(4π^2L^2 = \frac{ 500}{ 660}\)
\(L^2 = \frac{ 500}{ 660} \times \frac{1}{ 4π^2} =\frac{ 500}{ 660} \times \frac{1}{4 \times 10}\) ( since \(π^2\) = 10)
\(L^2 = 0.0189\)
∴ L = \(\sqrt{ 0.0189}\) = 0.138H = 138mH
Pregunta 7 Informe
a. Define each of the following terms used with simple machines:
i. Pivot ii.Load iii. Efficiency.
b. A truck of mass 1.2 × \(10^3\) kg is pulled from rest by a constant horizontal force of 25.2N on a leveled road. If the maximum speed attainable in the process is 60 km/h.
Calculate the: i. work done by the force; ii. distance traveled by the truck in reaching the maximum speed.
c. State two differences between absolute zero temperature and ice point.
d. An uncalibrated liquid-in-glass thermometer was used in determining a Celsius temperature. The readings are tabulated below
| Temperature/°C | -6 | 0 | 100 |
| Length of column/ cm | L | 2.0 | 15.0 |
a
i. A pivot, often referred to as a fulcrum, is the point or axis around which a simple machine, such as a lever or a seesaw, rotates.
ii. A load is the object or weight that is being moved, lifted, or supported by the machine.
iii. Efficiency refers to how effectively a machine can perform its intended task while minimizing energy loss
b.
i. Given: F = 25.2N, m = 1200kg, u = 0m/s , v = 60km/h = 16.67m/s
Workdone = ΔK.E = \(K.E_2 - K. E_1\)
Workdone = \(\frac{1}{2}K.E_2 - \frac{1}{2}K.E_1\)
Workdone = \(\frac{1}{2}m(v^2 - u^2)\)
Workdonk = \(\frac{1}{2}\times 1200 \times 16.67^2\) ( since u = 0)
Workdone = 600 x 277.89 = 166733.34J
Therefore, Workdone = \(1.67 \times 10^5\)J
ii. F = ma → a = \(\frac{F}{m} = \frac{25.2}{1.2 \times10^3} = 0.021ms^2\)
recall, \(v^2 = u^2 +2aS\)
S = \(\frac{v^2 - u^2}{ 2a} = \frac{16.67^2 - 0^2}{ 2\times 0.021}\)
S = 6616.4m = 6.62km.
c. Absolute zero temperature is the lowest possible temperature that theoretically represents the complete absence of thermal energy while the ice point is the temperature at which water coexists with ice in thermal equilibrium.
- Absolute zero temperature is defined as 0 Kelvin (0 K) or approximately -273.15°C while ice point is typically defined as 0°C at sea level
d. \(\frac{0 - ( - 6)}{ 2 - L} = \frac{ 100 - 0 }{ 15 - 2}\)
\(\frac{ 6}{ 2 - L} = \frac{100 }{ 13}\)
6 x 13 = 100( 2 - L )
78 = 200 - 100L
100L = 200 - 78
100L = 122
L = \(\frac{122}{100}\) = 1.22cm
Detalles de la respuesta
a
i. A pivot, often referred to as a fulcrum, is the point or axis around which a simple machine, such as a lever or a seesaw, rotates.
ii. A load is the object or weight that is being moved, lifted, or supported by the machine.
iii. Efficiency refers to how effectively a machine can perform its intended task while minimizing energy loss
b.
i. Given: F = 25.2N, m = 1200kg, u = 0m/s , v = 60km/h = 16.67m/s
Workdone = ΔK.E = \(K.E_2 - K. E_1\)
Workdone = \(\frac{1}{2}K.E_2 - \frac{1}{2}K.E_1\)
Workdone = \(\frac{1}{2}m(v^2 - u^2)\)
Workdonk = \(\frac{1}{2}\times 1200 \times 16.67^2\) ( since u = 0)
Workdone = 600 x 277.89 = 166733.34J
Therefore, Workdone = \(1.67 \times 10^5\)J
ii. F = ma → a = \(\frac{F}{m} = \frac{25.2}{1.2 \times10^3} = 0.021ms^2\)
recall, \(v^2 = u^2 +2aS\)
S = \(\frac{v^2 - u^2}{ 2a} = \frac{16.67^2 - 0^2}{ 2\times 0.021}\)
S = 6616.4m = 6.62km.
c. Absolute zero temperature is the lowest possible temperature that theoretically represents the complete absence of thermal energy while the ice point is the temperature at which water coexists with ice in thermal equilibrium.
- Absolute zero temperature is defined as 0 Kelvin (0 K) or approximately -273.15°C while ice point is typically defined as 0°C at sea level
d. \(\frac{0 - ( - 6)}{ 2 - L} = \frac{ 100 - 0 }{ 15 - 2}\)
\(\frac{ 6}{ 2 - L} = \frac{100 }{ 13}\)
6 x 13 = 100( 2 - L )
78 = 200 - 100L
100L = 200 - 78
100L = 122
L = \(\frac{122}{100}\) = 1.22cm
Pregunta 8 Informe
a. A projectile is fired at an angle, θ, to the horizontal with velocity, u. Show that at any time, t, during the motion, the: i. horizontal component of the velocity is independent of t;
ii. vertical component of the velocity depends on t.
b. State the assumption on which projectile motion is based.
i. From v = u + gt:
\(u_x\) = ucosθ, g = 0 (horizontal component is not affected by gravity)
\(v_x\) = ucosθ + (0)t
∴ \(v_x\) = ucosθ
ii. \(v_y = u_y + gt\)
∴ \(v_y = usinθ + gt\)
b. It assumes that the only force acting on the projectile is gravity, and there are no other forces, such as air resistance or drag, affecting its motion.
Detalles de la respuesta
i. From v = u + gt:
\(u_x\) = ucosθ, g = 0 (horizontal component is not affected by gravity)
\(v_x\) = ucosθ + (0)t
∴ \(v_x\) = ucosθ
ii. \(v_y = u_y + gt\)
∴ \(v_y = usinθ + gt\)
b. It assumes that the only force acting on the projectile is gravity, and there are no other forces, such as air resistance or drag, affecting its motion.
