QUESTION 5
Question 5.1
Newton's Universal Gravitational Law states that every point mass attracts every other point mass in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Question 5.2
To show that the mass of Mars is \(5.98 \times 10^{23} , \text{kg}\) and the radius is \(3,190,000 , \text{m}\):
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Mass of Mars: Given that the mass of Mars is about 10% that of the Earth: \[ M_{Mars} = 0.1 \times M_{Earth} \] The mass of Earth is approximately \(5.98 \times 10^{24} , \text{kg}\), hence: \[ M_{Mars} = 0.1 \times 5.98 \times 10^{24} = 5.98 \times 10^{23} , \text{kg} \]
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Radius of Mars: The radius of Mars is 2 times smaller than that of Earth. Assuming Earth's average radius is approximately \(6,371,000 , \text{m}\): \[ R_{Mars} = \frac{R_{Earth}}{2} = \frac{6,371,000}{2} = 3,185,500 \approx 3,190,000 , \text{m} , (\text{rounded to the nearest meter}) \]
Question 5.3
To calculate the mass of the magic school bus, we can use the relationship between weight (force due to gravity), mass, and gravitational acceleration.
Weight \(W = mg\) where:
- \(W\) = weight in Newtons
- \(m\) = mass in kg
- \(g\) = acceleration due to gravity on Mars
Using the weight of the bus on Mars: \[ 28,567 , \text{N} = m \cdot g_{Mars} \]
Now we need to find \(g_{Mars}\). From previous calculations, \(g_{Mars} = \frac{G \cdot M_{Mars}}{R_{Mars}^2}\). From data, let's assume \(g_{Mars} \approx 3.71 , \text{m/s}^2\).
Thus: \[ 28,567 , \text{N} = m \cdot 3.71 , \text{m/s}^2 \]
Now, we can solve for \(m\): \[ m = \frac{28,567 , \text{N}}{3.71 , \text{m/s}^2} \approx 7,700 , \text{kg} \]
Question 5.4
The gravitational acceleration on Mars can be calculated using the formula:
\[ g_{Mars} = \frac{G \cdot M_{Mars}}{R_{Mars}^2} \]
Where \(G \approx 6.674 \times 10^{-11} , \text{N m}^2/\text{kg}^2\), \[ M_{Mars} = 5.98 \times 10^{23} , \text{kg} \] \[ R_{Mars} = 3,190,000 , \text{m} \] By plugging into the formula: \[ g_{Mars} = \frac{6.674 \times 10^{-11} \cdot 5.98 \times 10^{23}}{(3,190,000)^2} \] Calculating gives: \[ g_{Mars} \approx 3.71 , \text{m/s}^2 \]
Question 5.5
STAYS THE SAME. The force of gravitational attraction depends on the mass of the planet and the distance from its center to the object. Since the mass of the bus increases, but it is still on the same planet (Mars), the force of gravitational attraction (which is directly caused by the mass of Mars and the radius) does not change due to the addition of the mass of the passengers.
QUESTION 6
Question 6.1.1
Angle of incidence is defined as the angle formed between the incident ray and the normal line (an imaginary line perpendicular to the surface at the point of incidence) when light hits a surface.
Question 6.1.2
To calculate the refractive index \( n \) for material X, we use Snell's Law:
\[ n = \frac{\sin(\text{angle of incidence})}{\sin(\text{angle of refraction})} \]
If given angles are, for instance, \( i \) as 30° and \( r \) as 20°, Then: \[ n_X = \frac{\sin(30)}{\sin(20)} \] Calculating will give the refractive index.
Question 6.1.3
The speed of light in a material can be calculated using the formula:
\[ v = \frac{c}{n} \]
Where:
- \( c \) is the speed of light in a vacuum (\( c \approx 3.00 \times 10^8 , \text{m/s} \))
Plug in the calculated refractive index to find \( v \).
Question 6.2.1
Using Snell's Law again for material Y: \[ n_Y = \frac{\sin(7°)}{\sin(13°)} \]
Question 6.2.2
It is important to consider the refractive indices of possible contact lenses because:
- Reflection: A high refractive index may reflect more light, leading to reduced visibility.
- Refraction: Correct matching of refractive indices ensures that light focuses correctly on the retina for clear vision.
QUESTION 7
Question 7.1.1
Diffraction is the bending and spreading of waves around obstacles and apertures, which occurs when a wave encounters a slit or an edge.
Question 7.1.2
[Light intensity curve is required for this answer; it cannot be visually completed without graphical reference.]
Question 7.1.3
One safety precaution is to avoid direct eye exposure to the laser beam to prevent potential eye damage.
Question 7.1.4
Constructive interference is taking place in the middle of the central bright band due to the alignment of wavefronts leading to an increase in amplitude.
Question 7.1.5
Two changes that a person can make:
- Decrease the width of the slit – which increases diffraction.
- Increase the wavelength of the laser light used – which also increases the distance between bright and dark bands.
Question 7.2.1
The diffraction pattern produced by the blue laser would be the one with a narrower spread (or different intensity pattern), as blue light has a shorter wavelength.
Question 7.2.2
The blue light produces a more concentrated pattern due to its shorter wavelength compared to red light. Thus, it shows a sharper and more defined diffraction pattern.
1 answer