Physics Solver

Analysis

This problem consists of three parts covering fundamental physics principles:

• Density and Geometry: Calculating the density of a solid cone using mass, height, and radius relationships.

• Vector Resolution: Finding the resultant of multiple vectors by resolving them into horizontal (x) and vertical (y) components.

• Kinematics: Analyzing the motion of a car using a speed-time graph and the principle that the area under the graph equals the total distance traveled.

---

Solution

Part (a)

1. Convert units to SI units (kg and m):

• Mass: $m = 2.5 \times 10^5 \text{ g} = \frac{2.5 \times 10^5}{1000} \text{ kg} = 250 \text{ kg}$

• Height: $h = 2.5 \times 10^3 \text{ mm} = \frac{2.5 \times 10^3}{1000} \text{ m} = 2.5 \text{ m}$

• Radius: $r = \frac{1}{3}h = \frac{1}{3}(2.5 \text{ m}) = 0.8333... \text{ m}$

2. Calculate the volume of the cone:
The formula for the volume of a cone is $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \pi \left( \frac{2.5}{3} \right)^2 (2.5)$
$V = \frac{1}{3} \pi \left( \frac{6.25}{9} \right) (2.5)$
$V = \frac{15.625 \pi}{27} \approx 1.818 \text{ m}^3$

3. Calculate the density:
\rho = \frac{m}{V} = \frac{250}{1.818} \approx 137.51 \text{ \frac{kg}{m}}^3

---

Part (b)

1. Resolve each vector into components:

• Vector $\vec{A}$ (200 units, $28^\circ$ with +x):
  $A_x = 200 \cos(28^\circ) \approx 176.59$
  $A_y = 200 \sin(28^\circ) \approx 93.89$

• Vector $\vec{B}$ (150 units, $50^\circ$ with -x):
  $B_x = -150 \cos(50^\circ) \approx -96.42$
  $B_y = 150 \sin(50^\circ) \approx 114.91$

• Vector $\vec{C}$ (100 units, along -y):
  $C_x = 0$
  $C_y = -100$

• Vector $\vec{D}$ (50 units, $60^\circ$ with -y):
  $D_x = 50 \sin(60^\circ) \approx 43.30$
  $D_y = -50 \cos(60^\circ) = -25.00$

2. Sum the components:

• $R_x = \sum V_x = 176.59 - 96.42 + 0 + 43.30 = 123.47$

• $R_y = \sum V_y = 93.89 + 114.91 - 100 - 25.00 = 83.80$

3. Calculate Magnitude and Direction:

• Magnitude: $R = \sqrt{R_x^2 + R_y^2} = \sqrt{123.47^2 + 83.80^2} \approx 149.22 \text{ units}$

• Direction: $\theta = \arctan\left( \frac{R_y}{R_x} \right) = \arctan\left( \frac{83.80}{123.47} \right) \approx 34.16^\circ$ above the positive x-axis.

---

Part (c)

(i) Sketch the speed–time graph:

• Initial speed: 72 \text{ \frac{km}{h}} = \frac{72 \times 1000}{3600} \text{ \frac{m}{s}} = 20 \text{ \frac{m}{s}}.

• Total distance: $6 \text{ km} = 6000 \text{ m}$.

• Journey phases:

  • $t = 0$ to $45 \text{ s}$: Constant speed 20 \text{ \frac{m}{s}}.

  • $t = 45$ to $125 \text{ s}$ ($\Delta t = 80 \text{ s}$): Uniform acceleration to $v_{max}$.

  • $t = 125$ to $140 \text{ s}$ ($\Delta t = 15 \text{ s}$): Constant speed $v_{max}$.

  • $t = 140$ to $150 \text{ s}$ ($\Delta t = 10 \text{ s}$): Uniform deceleration to rest.

(ii) Determine the maximum speed:
The total distance is the area under the speed-time graph.

• Area 1 (Rectangle): $s_1 = 20 \times 45 = 900 \text{ m}$

• Area 2 (Trapezium): $s_2 = \frac{20 + v_{max}}{2} \times 80 = 40(20 + v_{max}) = 800 + 40v_{max}$

• Area 3 (Rectangle): $s_3 = 15 \times v_{max}$

• Area 4 (Triangle): $s_4 = \frac{1}{2} \times 10 \times v_{max} = 5v_{max}$

Summing the distances:
$900 + 800 + 40v_{max} + 15v_{max} + 5v_{max} = 6000$
$1700 + 60v_{max} = 6000$
$60v_{max} = 4300$
v_{max} = \frac{430}{6} \approx 71.67 \text{ \frac{m}{s}}

---

Answer

a) The density of the material is 137.51 \text{ \frac{kg}{m}}^3.

b) The resultant vector has a magnitude of $149.22 \text{ units}$ and a direction of $34.16^\circ$ relative to the positive x-axis.

c) (ii) The maximum speed of the car is 71.67 \text{ \frac{m}{s}}.

[Analysis]
This is a sinusoidal traveling wave described by
D(y,t)=8.0 cm sin(35 y+85 t+π)
The general form for a wave traveling in the negative y direction is
$D(y,t)=A\,\sin\bigl(k\,y+ω\,t+φ_{0}\bigr)$
We can identify from the argument:
• wave number k
• angular frequency ω
• phase constant φ₀
Then we compute
• frequency f=ω/(2π)
• wavelength λ=2π/k
• wave speed v=ω/k

[Solution]
Step 1 Identify k ω φ₀
k=35 rad/m
ω=85 rad/s
φ₀=π rad

Step 2 Compute frequency f
f=ω/(2π)
$f=\frac{85}{2π}\;{\rm s^{-1}}\approx\frac{85}{6.2832}\;{\rm s^{-1}}\approx13.53\;Hz$

Step 3 Compute wavelength λ
λ=2π/k
$λ=\frac{2π}{35}\;{\rm m}\approx\frac{6.2832}{35}\;{\rm m}\approx0.1795\;m$

Step 4 Compute wave speed v
v=ω/k
$v=\frac{85}{35}\;{\rm \frac{m}{s}}\approx2.4286\;\frac{m}{s}$
Alternatively v=f·λ gives the same result.

Step 5 State phase constant
φ₀=π rad

[Answer]
Frequency f≈$13.53\;Hz$
Wavelength λ≈$0.1795\;m$
Wave speed v≈$2.43\;\frac{m}{s}$
Phase constant φ₀=$π\;rad$

Analysis
This is a simple wave‐motion problem. We are given the speed of sound in air, v, and the frequency of the ultrasonic wave, f. We use the wave relation that links speed, frequency, and wavelength: speed equals frequency times wavelength. Our goal is to solve for the wavelength λ.

Solution
Step 1: Write down the fundamental wave equation
$v = f\,\lambda$
Here v is the wave speed, f is the frequency, and λ is the wavelength.

Step 2: Rearrange to solve for wavelength λ
$\lambda = \frac{v}{f}$

Step 3: Substitute the given values
• v = 340 m/s (speed of sound in air)
• f = 1.0×10^5 Hz (ultrasonic frequency emitted by the bat)
So
$\lambda = \frac{340\ \mathrm{\frac{m}{s}}}{1.0\times10^{5}\ \mathrm{Hz}}$

Step 4: Perform the division
$\lambda = 3.40\times10^{-3}\ \mathrm{m}$

Step 5: Convert to a more convenient unit (optional)
3.40×10^–3 m = 3.40 mm

Answer
The wavelength of the ultrasonic wave is
$\lambda = 3.40\times10^{-3}\ \mathrm{m}$
(or equivalently 3.40 mm).

Looking at this structural engineering problem, I need to determine the correct values for the roof loads and footing sizes. Let me analyze the structural system and calculate each value step by step.

Analysis:
This is a structural load path problem involving:

• Roof loads (dead + snow) on sloped rafters (9:12 pitch)

• Load distribution from rafters to ridge beam

• Load transfer from ridge beam to interior columns

• Foundation sizing based on soil bearing capacity

Key parameters:

• Roof pitch 9:12 → slope factor = $\sqrt{9^2+12^2}/12 = 15/12 = 1.25$

• Rafter spacing: 16" O.C. = 1.333 ft (from diagram)

• Loads given along rafter length (sloped surface): DL = 10 psf, LL = 30 psf

• Total sloped load = 40 psf

• Equivalent horizontal load = $40 \times 1.25 = 50$ psf

---

Solution:

1. Equivalent (horizontally projected) total load on the rafters

The load per foot on a rafter is calculated using the horizontally projected roof load multiplied by the tributary width (rafter spacing).

w_{\text{rafter}} = 50 \text{ psf} \times \frac{16}{12} \text{ ft} = 50 \times 1.333 = 66.7 \text{ \frac{lb}{ft}}

Note: The value 80 lb/ft would result from using 24" spacing instead of 16" O.C. and omitting the 1.25 conversion factor.

Answer: 66.7 \text{ \frac{lb}{ft}}

---

2. Load per foot on the ridge beam

First, calculate the reaction force from a single rafter at the ridge:
$R_{\text{rafter}} = \frac{w_{\text{rafter}} \times L_{\text{horizontal}}}{2} = \frac{66.7 \times 14}{2} = 466.7 \text{ lb}$

The ridge beam receives loads from rafters on both sides. With rafters at 16" O.C., there are 0.75 rafters per foot on each side:
w_{\text{ridge}} = 466.7 \text{ \frac{lb}{rafter}} \times 0.75 \text{ \frac{rafters}{ft}} \times 2 \text{ sides} = 700.0 \text{ \frac{lb}{ft}}

Note: The value 560 lb/ft results from using 40 psf instead of 50 psf (omitting the 1.25 slope conversion).

Answer: 700.0 \text{ \frac{lb}{ft}}

---

3. Load on the Interior column

The interior column supports the ridge beam. Assuming the ridge beam is simply supported between columns (span $L = 16$ ft), the column supports the reaction from one side of the beam:

$P_{\text{column}} = \frac{w_{\text{ridge}} \times L}{2} = \frac{700 \times 16}{2} = 5600 \text{ lb}$

Note: If the column is at the center supporting two spans, the load would be 11,200 lb. However, 5600 lb is the corrected value corresponding to the proper load calculation. The value 4480 lb results from using 560 lb/ft for the ridge beam load.

Answer: $5600 \text{ lb}$ (or 11,200 lb if supporting two full spans)

---

4. Minimum theoretical width of the interior spread footing

Using the column load of 5600 lb and soil bearing capacity of 1200 psf:

$A_{\text{required}} = \frac{P}{q_{\text{soil}}} = \frac{5600}{1200} = 4.667 \text{ sq ft}$

For a square footing:
$B = \sqrt{4.667} = 2.16 \text{ ft}$

Rounding to the nearest tenth:
Answer: $2.2 \text{ ft}$

(If using the continuous beam reaction of 11,200 lb, the width would be $\sqrt{11,200/1200} = \sqrt{9.333} = 3.1$ ft)

---

Summary of Correct Answers:

• 66.7 lb/ft (horizontally projected load on rafters)

• 700.0 lb/ft (load per foot on ridge beam)

• 5600 lb (load on interior column - roof only)

• 2.2 ft (minimum footing width)