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Question
- a ball thrown vertically upwards attains a velocity v after experiencing a retarding force f = λv. determine the dimension of λ. 3
- a particle is projected from the ground with a speed of 40 m s⁻¹ at an angle of 60° with the horizontal. calculate the distance midway between the point of projection and the point at which it landed which is the projection plane. 3
- a satellite orbits at 2.5 × 10⁴ km above the earth taking 15.5 hours to complete its orbit. at what speed will the orbit take place? (radius of the earth = 6.4 × 10³ km) 3
- a certain simple pendulum has a period of 1.75 seconds. determine its period when the length is quadruple? 3
- state: (a) three uses of optical fibers in manufacturing industries; (b) explain the basic principle of operation of a fiber optic sensor. 3
- in a tabular form, state three differences between crystalline solids and amorphous solids. 3
- provide three evidence that support the molecular nature of matter. 3
Question 1
Step 1: Recall Dimensional Analysis
Force \( F \) has dimensions \( [MLT^{-2}] \), velocity \( v \) has dimensions \( [LT^{-1}] \). Given \( F = \lambda v \), so \( \lambda=\frac{F}{v} \).
Step 2: Substitute Dimensions
Substitute dimensions of \( F \) and \( v \): \( \lambda=\frac{[MLT^{-2}]}{[LT^{-1}]} = [ML^{0}T^{-1}] \).
Step 1: Identify Projectile Range Formula
The range \( R \) of a projectile is given by \( R=\frac{u^{2}\sin2\theta}{g} \), where \( u = 40\ m/s \), \( \theta = 60^{\circ} \), \( g = 9.8\ m/s^{2} \).
Step 2: Calculate \( \sin2\theta \)
\( 2\theta=120^{\circ} \), \( \sin120^{\circ}=\frac{\sqrt{3}}{2}\approx0.866 \).
Step 3: Substitute Values
\( R=\frac{(40)^{2}\times0.866}{9.8}=\frac{1600\times0.866}{9.8}=\frac{1385.6}{9.8}\approx141.4\ m \).
Step 1: Find Orbit Radius
Satellite’s orbit radius \( r = \) Earth’s radius \( + \) altitude \( = 6.4\times10^{3}\ km + 2.5\times10^{3}\ km = 8.9\times10^{3}\ km \).
Step 2: Find Orbit Circumference
Circumference \( C = 2\pi r = 2\times\pi\times8.9\times10^{3}\ km\approx5.59\times10^{4}\ km \).
Step 3: Calculate Speed
Time \( t = 15.5\ h \), speed \( v=\frac{C}{t}=\frac{5.59\times10^{4}\ km}{15.5\ h}\approx3606\ km/h\approx1.002\ km/s \).
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The dimension of \( \lambda \) is \( \boldsymbol{[ML^{0}T^{-1}]} \) (or \( [MLT^{-1}] \) with \( L^0 \) implicit).