WAEC SSCE - Physics - 2023 (Essay)
Tambaya 1 Rahoto
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
Bayanin Amsa
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
Tambaya 2 Rahoto
The force, F, acting on the wings of an aircraft moving through the air of velocity, v, and density, ρ, is given by the equation F = \(kv^xρ^yA^z\), where k is a dimensionless constant and A is the surface area of the wings of the aircraft. Use dimensional analysis to determine the values of x, y, and z.
F = \(kv^xρ^yA^z\)
For the left-hand side:
F = mass × acceleration = \(MLT^{-2}\)
For the right-hand side:
v = \(LT^{-1}\), ρ = \(ML^{-3}\) and A = \(L^2\)
So,
\(MLT^{-2} = k(LT^{-1})^x(ML^{-3})^y(L^2)^z\)
Since k is dimensionless,
\(MLT^{-2} = (LT^{-1})^x(ML^{-3})^y(L^2)^z\)
\(MLT^{-2} = L^x × T^{(-1)x} \times M^y × L^{(-3)y} \times L^{(2)z}\)
\(MLT^{-2} = L^x × T^{-x} \times M^y \times L^{-3y} \times L^{2z}\)
\(MLT^{-2} = L^{(x - 3y + 2z)} \times T^{-x} \times M^y\)
\(M^1L^1T^{-2}= L^{(x - 3y + 2z)} \times T^{-x} \times M^{y}\)
Comparing the powers:
For M,
y = 1
For L,
x - 3y + 2z = 1 ---- (i)
For T,
-x = -2
∴ x = 2
Substitute (2) for x and (1) for y in equation (i)
⇒ 2 - 3(1) + 2z = 1
⇒ 2 - 3 + 2z = 1
⇒ -1 + 2z = 1
⇒ 2z = 1 + 1
⇒ 2z = 2
∴ z = 1
Hence, x = 2, y = 1 and z = 1.
Bayanin Amsa
F = \(kv^xρ^yA^z\)
For the left-hand side:
F = mass × acceleration = \(MLT^{-2}\)
For the right-hand side:
v = \(LT^{-1}\), ρ = \(ML^{-3}\) and A = \(L^2\)
So,
\(MLT^{-2} = k(LT^{-1})^x(ML^{-3})^y(L^2)^z\)
Since k is dimensionless,
\(MLT^{-2} = (LT^{-1})^x(ML^{-3})^y(L^2)^z\)
\(MLT^{-2} = L^x × T^{(-1)x} \times M^y × L^{(-3)y} \times L^{(2)z}\)
\(MLT^{-2} = L^x × T^{-x} \times M^y \times L^{-3y} \times L^{2z}\)
\(MLT^{-2} = L^{(x - 3y + 2z)} \times T^{-x} \times M^y\)
\(M^1L^1T^{-2}= L^{(x - 3y + 2z)} \times T^{-x} \times M^{y}\)
Comparing the powers:
For M,
y = 1
For L,
x - 3y + 2z = 1 ---- (i)
For T,
-x = -2
∴ x = 2
Substitute (2) for x and (1) for y in equation (i)
⇒ 2 - 3(1) + 2z = 1
⇒ 2 - 3 + 2z = 1
⇒ -1 + 2z = 1
⇒ 2z = 1 + 1
⇒ 2z = 2
∴ z = 1
Hence, x = 2, y = 1 and z = 1.
Tambaya 3 Rahoto
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.
Bayanin Amsa
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.
Tambaya 4 Rahoto
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.
Bayanin Amsa
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.
Tambaya 5 Rahoto
a. Define strain energy.
b. Write an expression for the energy stored, E, in a stretched wire of original length, l , cross-sectional area, A, extension, e, and Young's modulus, Y, of the material of the wire.
a. Strain energy is the potential energy stored within an elastic material as a result of its deformation caused by external forces.
b. Energy stored E = \(\frac{1}{2}Fe\)
Young Modulus = \(\frac{ Stress}{ Strain}\) = Y
Stress = \(\frac{Force}{Area} = \frac{(F)}{(A)}\)
Strain = \(\frac{ Extension}{Original Length} = \frac{(e)}{(l)}\)
Y = \(\frac{F}{A}\div \frac{e}{l}\)
Y = \(\frac{F}{A} \times \frac{l}{e}\)
F = \(\frac{YAe}{l}\)
Therefore, Energy stored E = \(\frac{1}{2}\frac{YAe}{l}\times e\)
E = \(\frac{YAe^2}{2l}\)
Bayanin Amsa
a. Strain energy is the potential energy stored within an elastic material as a result of its deformation caused by external forces.
b. Energy stored E = \(\frac{1}{2}Fe\)
Young Modulus = \(\frac{ Stress}{ Strain}\) = Y
Stress = \(\frac{Force}{Area} = \frac{(F)}{(A)}\)
Strain = \(\frac{ Extension}{Original Length} = \frac{(e)}{(l)}\)
Y = \(\frac{F}{A}\div \frac{e}{l}\)
Y = \(\frac{F}{A} \times \frac{l}{e}\)
F = \(\frac{YAe}{l}\)
Therefore, Energy stored E = \(\frac{1}{2}\frac{YAe}{l}\times e\)
E = \(\frac{YAe^2}{2l}\)
Tambaya 6 Rahoto
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.
Bayanin Amsa
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.
Tambaya 7 Rahoto
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
Bayanin Amsa
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
Tambaya 8 Rahoto
State three differences between geostationary satellites and polar satellites.
| Geostationary satellites | Polar satellites | |
| 1 | They orbit the earth directly above the equator in geosynchronous orbit. they remain fixed relative to a specific point on the Earth's surface | They orbit the Earth, passing over or near the north and south poles on each orbit. |
| 2 | They have an inclination of 0°, meaning their orbital plane aligns with the equatorial plane | They have a high inclination angle( usually around 90°) to achieve polar orbits. |
| 3 | They observe the Earth from a fixed position, providing a continuous view of a specific region. | They observe the Earth from a changing perspective as the orbit, resulting in different views of the Earth's surface with each pass |
| 4 | They primarily collect data for meteorological observation weather forecasting, and telecommunications. | They collect data for a wide range of applications, including weather monitoring, climate research, Environmental monitoring, and scientific research. |
| 5 | They are positioned at an altitude of approximately 35,786 kilometres above the Earth's surface. | They operate at a lower altitudes, typically between 700 and 1500 kilometers |
| 6 | They provide a continuous stream of data and observations for a specific region, allowing for real-time monitoring. | They have a lower data refresh rate since they pass over a given less frequently. However, they provide a broader coverage. |
pick any three according to the given instructions.
Bayanin Amsa
| Geostationary satellites | Polar satellites | |
| 1 | They orbit the earth directly above the equator in geosynchronous orbit. they remain fixed relative to a specific point on the Earth's surface | They orbit the Earth, passing over or near the north and south poles on each orbit. |
| 2 | They have an inclination of 0°, meaning their orbital plane aligns with the equatorial plane | They have a high inclination angle( usually around 90°) to achieve polar orbits. |
| 3 | They observe the Earth from a fixed position, providing a continuous view of a specific region. | They observe the Earth from a changing perspective as the orbit, resulting in different views of the Earth's surface with each pass |
| 4 | They primarily collect data for meteorological observation weather forecasting, and telecommunications. | They collect data for a wide range of applications, including weather monitoring, climate research, Environmental monitoring, and scientific research. |
| 5 | They are positioned at an altitude of approximately 35,786 kilometres above the Earth's surface. | They operate at a lower altitudes, typically between 700 and 1500 kilometers |
| 6 | They provide a continuous stream of data and observations for a specific region, allowing for real-time monitoring. | They have a lower data refresh rate since they pass over a given less frequently. However, they provide a broader coverage. |
pick any three according to the given instructions.
