physics theory (kunle)
q1. the diagram below illustrates the trajectory of a fired missile from point p at 250ms⁻¹
diagram: y-axis, x-axis, angle 42°, velocity 250ms⁻¹, trajectory from p to q
the missile hits point q after 40s, calculate the distance |pq|
q2. state three properties of an intrinsic semiconductor.
q3. a 50n force is applied to the free end of a spiral spring of force constant 100nm⁻¹. calculate the work done by the force to stretch the spring.
q4. what is a geostationary satellite.
q5. state the s.i unit of the intensity of a blackbody radiation.
5b. state two features of the intensity-wavelength graph of a perfect blackbody at different temperature.

Question

Accepted Answer

### Q1 Solution:
# Explanation:
## Step1: Identify horizontal component
The horizontal component of velocity $ v_x = v \cos\theta $, where $ v = 250 \, \text{m/s} $, $ \theta = 42^\circ $. So $ v_x = 250 \cos(42^\circ) $.
## Step2: Calculate distance $ |PQ| $
Distance $ d = v_x \times t $, $ t = 40 \, \text{s} $. First, $ \cos(42^\circ) \approx 0.7431 $, so $ v_x \approx 250 \times 0.7431 = 185.775 \, \text{m/s} $. Then $ d = 185.775 \times 40 = 7431 \, \text{m} $.
# Answer:
$ 7431 \, \text{m} $ (or approximately $ 7.43 \times 10^3 \, \text{m} $)

### Q2 Solution:
# Brief Explanations:
1. Intrinsic semiconductors have equal number of electrons and holes at a given temperature (in pure state).
2. Their conductivity increases with increase in temperature (as more electron - hole pairs are generated).
3. At absolute zero temperature, they behave as insulators (no free charge carriers).
# Answer:
1. Equal number of electrons and holes (in pure state).
2. Conductivity increases with temperature.
3. Behaves as an insulator at 0 K.

### Q3 Solution:
# Explanation:
## Step1: Recall Hooke's Law and work formula
Hooke's Law: $ F = kx $, work done on a spring $ W=\frac{1}{2}kx^2 $. First, find $ x $ from $ F = kx $, so $ x=\frac{F}{k} $.
## Step2: Substitute values
Given $ F = 50 \, \text{N} $, $ k = 100 \, \text{N/m} $, so $ x=\frac{50}{100}=0.5 \, \text{m} $. Then $ W=\frac{1}{2}\times100\times(0.5)^2 $.
## Step3: Calculate work
$ W=\frac{1}{2}\times100\times0.25 = 12.5 \, \text{J} $.
# Answer:
$ 12.5 \, \text{J} $

### Q4 Solution:
# Brief Explanations:
A geostationary satellite is a satellite that orbits the Earth with a period equal to the Earth's rotational period (24 hours), has an orbital radius of approximately $ 4.23\times 10^{4} \, \text{km} $ (or $ 3.58\times 10^{7} \, \text{m} $) above the Earth's equator, and appears stationary relative to a point on the Earth's surface (since its angular velocity matches the Earth's angular velocity of rotation).
# Answer:
A geostationary satellite is a satellite that orbits the Earth with a period of 24 hours (matching Earth's rotation period), is located above the Earth's equator, and appears stationary relative to a point on the Earth's surface.

### Q5 Solution:
# Brief Explanations:
The SI unit of the intensity of black - body radiation is the watt per square meter ($ \text{W/m}^2 $). Intensity is power per unit area, and the SI unit of power is watt (W) and area is square meter ($ \text{m}^2 $).
# Answer:
$ \text{W/m}^2 $ (watt per square meter)

### Q5b Solution:
# Brief Explanations:
1. The intensity - wavelength graph of a perfect black - body has a peak (maximum intensity) at a certain wavelength, and this peak wavelength shifts towards shorter wavelengths as the temperature of the black - body increases (Wien's displacement law).
2. The total intensity (area under the curve) of the black - body radiation increases with the increase in temperature (Stefan - Boltzmann law, as total radiated power is proportional to $ T^4 $, and intensity is power per unit area).
# Answer:
1. The peak wavelength (where intensity is maximum) decreases as temperature increases.
2. The total intensity (area under the curve) increases with increasing temperature.

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QUESTION IMAGE

Question

Explanation:

Q1 Solution:

Step1: Identify horizontal component

Step2: Calculate distance \( |PQ| \)

Step1: Recall Hooke's Law and work formula

Step2: Substitute values

Step3: Calculate work

Answer:

Q2 Solution: