Convert 2.5 g/mm2 to kg/m2 and perform dimensional analysis on T = 2π sqrt(m/Fx).

$1 \text{ g} = 10^{-3} \text{ kg}$
So, $2.5 \text{ g} = 2.5 \times 10^{-3} \text{ kg}$.

$1 \text{ mm} = 10^{-3} \text{ m}$
$1 \text{ mm}^2 = (10^{-3} \text{ m})^2 = 10^{-6} \text{ m}^2$
Therefore, $1 \text{ mm}^{-2} = (10^{-6} \text{ m}^2)^{-1} = 10^6 \text{ m}^{-2}$.

The converted value is:

Question 1. b: Dimensional analysis of $T = 2\pi \sqrt{\frac{m}{Fx}}$

• Time period $T$: $[T]$
• Mass $m$: $[M]$
• Force $F$: $[MLT^{-2}]$ (since $F = ma$)
• Distance $x$: $[L]$
• $2\pi$ is a dimensionless constant.

The term inside the square root is $\frac{m}{Fx}$.
Dimensions of $Fx$: $[MLT^{-2}] \times [L] = [ML^2T^{-2}]$
Dimensions of $\frac{m}{Fx}$: $\frac{[M]}{[ML^2T^{-2}]} = [M^{1-1}L^{-2}T^{2}] = [L^{-2}T^2]$
Dimensions of $\sqrt{\frac{m}{Fx}}$: $\sqrt{[L^{-2}T^2]} = [L^{-1}T]$

LHS dimensions: $[T]$
RHS dimensions: $[L^{-1}T]$
Since $[T] \neq [L^{-1}T]$, the dimensions are not equal.

The formula is:

Question 2. a: Conditions for two vectors to be equal and examples of "magnitude"

• Two vectors are considered equal if they have the same magnitude and the same direction.

• Examples of the term "magnitude" as applied to vector and scalar quantities:

  • The length of a displacement vector.
  • The speed of a velocity vector.
  • The mass of an object (a scalar quantity).
  • The temperature of a substance (a scalar quantity).

Question 2. b: Magnitude of the resultant of two equal vectors

If two vectors $\vec{A}$ and $\vec{B}$ have magnitudes $A$ and $B$ respectively, and the angle between them is $\theta$, the magnitude of their resultant $\vec{R}$ is given by:

Given that the two vectors have equal magnitudes, $A = F$ and $B = F$.

The magnitude of the resultant is:

Question 3. a: Total mechanical energy of a particle in SHM

Mass $m = 40 \text{ g} = 0.040 \text{ kg}$.
Amplitude $A = 2.0 \text{ cm} = 0.020 \text{ m}$.
Time period $T = 0.20 \text{ s}$.

The relationship between time period and angular frequency is $\omega = \frac{2\pi}{T}$.

For SHM, the angular frequency is also given by $\omega = \sqrt{\frac{k}{m}}$.
Squaring both sides: $\omega^2 = \frac{k}{m}$.
So, $k = m\omega^2$.

Using $\pi^2 \approx 9.87$:

The total mechanical energy of a system in SHM is given by $E = \frac{1}{2} k A^2$.

Using $\pi^2 \approx 9.87$:

Rounding to two significant figures:

The total mechanical energy of the system is:

Question 4. a: Tensile stress developed in the wire

Mass $m = 4.0 \text{ kg}$.
Radius $r = 2.0 \text{ mm} = 2.0 \times 10^{-3} \text{ m}$.
Acceleration due to gravity $g = 9.8 \text{ m/s}^2$.

At equilibrium, the tension force $F$ in the wire is equal to the weight of the load.

The area $A$ of a circular wire is $A = \pi r^2$.

Using $\pi \approx 3.14159$:

Tensile stress is defined as force per unit area: $\sigma = \frac{F}{A}$.

Rounding to two significant figures:

The tensile stress developed in the wire is:

Question 5. a: What is a fluid? Give two examples.

• A fluid is a substance that continuously deforms (flows) under an applied shear stress. Unlike solids, fluids cannot withstand shear stress at rest. They include liquids and gases.
• Examples: Water (liquid), Air (gas).

Question 5. b: Define the terms adhesion and cohesion.

• Adhesion: The attractive force between molecules of different substances. For example, water molecules adhering to glass.
• Cohesion: The attractive force between molecules of the same substance. For example, water molecules sticking to each other, forming droplets.

Question 6: Wave properties from $y = 5 \sin(x - 60t)$

The general equation for a sinusoidal wave traveling in the positive x-direction is $y = A \sin(kx - \omega t)$, where:

• $A$ is the amplitude
• $k$ is the wave number
• $\omega$ is the angular frequency Comparing $y = 5 \sin(x - 60t)$ with the general form:
• Amplitude $A = 5$ (units not specified, assume arbitrary units)
• Wave number $k = 1$ (units are $\text{rad/m}$ or $\text{m}^{-1}$)
• Angular frequency $\omega = 60$ (units are $\text{rad/s}$)

From comparison, $k = 1$.

The angular frequency $\omega$ is related to the frequency $f$ by $\omega = 2\pi f$.

The wave number $k$ is related to the wavelength $\lambda$ by $k = \frac{2\pi}{\lambda}$.

The velocity of the wave is given by $v = f\lambda$ or $v = \frac{\omega}{k}$.
Using $v = \frac{\omega}{k}$:

The wave number, frequency, wavelength, and velocity are:

• Wave number: $\boxed{1 \text{ m}^{-1}}$
• Frequency: $\boxed{\frac{30}{\pi} \text{ Hz}}$
• Wavelength: $\boxed{2\pi \text{ m}}$
• Velocity: $\boxed{60 \text{ m/s}}$