1. A student practicing for a track meet ran 263 m in 30 sec. What was her average speed?

Under the assumption that the beam is a rectangular cantilever beam that is free to vibrate, the theoretical first natural frequ

BartSMP [9]

Answer:

a) Δf = 0.7 n , e) f = (15.1 ± 0.7) 10³ Hz

Explanation:

This is an error about the uncertainty or error in the calculated quantities.

Let's work all the magnitudes is the SI system

The frequency of oscillation is

f = n / 2π L² √( E /ρ)

where n is an integer

Let's calculate the magnitude of the oscillation

f = n / 2π (0.2335)² √ (210 10⁹/7800)

f = n /0.34257 √ (26.923 10⁶)

f = n /0.34257 5.1887 10³

f = 15.1464 10³ n

a) We are asked for the uncertainty of the frequency (Df)

Δf = | df / dL | ΔL + df /dE ΔE + df /dρ Δρ

in this case no error is indicated in Young's modulus and density, so we will consider them exact

ΔE = Δρ = 0

Δf = df /dL ΔL

df = n / 2π √E /ρ | -2 / L³ | ΔL

df = n / 2π 5.1887 10³ | 2 / 0.2335³) 0.005 10⁻³

df = n 0.649

Absolute deviations must be given with a single significant figure

Δf = 0.7 n

b, c) The uncertainty with the width and thickness of the canteliver is associated with the density

In your expression there is no specific dependency so the uncertainty should be zero

The exact equation for the natural nodes is

f = n / 2π L² √ (E e /ρA)

where A is the area of the cantilever and its thickness,

In this case, they must perform the derivatives, calculate and approximate a significant figure

Δf = | df / dL | ΔL + df /de Δe + df /dA ΔA

Δf = 0.7 n + n 2π L² √(E/ρ A) | ½ 1/√e | Δe

+ n / 2π L² √(Ee /ρ) | 3/2 1√A23 |

the area is

A = b h

A = 24.9 3.3 10⁻⁶

A = 82.17 10⁻⁶ m²

DA = dA /db ΔB + dA /dh Δh

dA = h Δb + b Δh

dA = 3.3 10⁻³ 0.005 10⁻³ + 24.9 10⁻³ 0.005 10⁻³

dA = (3.3 + 24.9) 0.005 10⁻⁶

dA = 1.4 10⁻⁷ m²

let's calculate each term

A ’= n / 2π L² √a (E/ρ A) | ½ 1 /√ e | Δe

A ’= n/ 2π L² √ (E /ρ) | ½ 1 / (√e/√ A) |Δe

A ’= 15.1464 10³ n ½ 1 / [√ (24.9 10⁻³)/ √ (81.17 10⁻⁶)] 0.005 10⁻³

A '= 0.0266 n

A ’= 2.66 10⁻² n

A ’’ = n / 2π L² √ (E e /ρ) | 3/2 1 /√A³ |

A ’’ = n / 2π L² √(E /ρ) √ e | 3/2 1 /√ A³ | ΔA

A ’’ = n 15.1464 10³ 3/2 √ (24.9 10⁻³) /√ (82.17 10⁻⁶) 3 1.4 10⁻⁷

A ’’ = n 15.1464 1.5 1.5779 / 744.85 1.4 10⁴

A ’’ = 6,738 10²

we write the equation of uncertainty

Δf = n (0.649 + 2.66 10⁻² + 6.738 10²)

The uncertainty due to thickness is

Δf = 3 10⁻² n

The uncertainty regarding the area, note that this magnitude should be measured with much greater precision, specifically the height since the errors of the width are very small

Δf = 7 10² n

d) Δf = 7 10² n

e) the natural frequency n = 1

f = (15.1 ± 0.7) 10³ Hz

7 0

3 years ago

How do your results from ray tracing compare to your results from using the thin-lens equation?

EastWind [94]

Answer:

20cm

Explanation:

A convex lens has a positive focal length and the object placed in front of it produce both virtual and real image <em>(image distance can be negative or positive depending on the nature of the image</em>).

According to the lens equation

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ where;

f is the focal length of the lens

u is the object distance

v is the image distance

If the magnification is - 0.6

mag = v/u = -0.5

v = -0.5u

since v = 10cm

10 = -0.5u

u = -10/0.5

u =-20 cm

Substitute u = -20cm ( due to negative magnification)and v = 10cm into the lens formula to get the focal length f

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{-1+2}{20} \\\frac{1}{f} = \frac{1}{20} \\cross \ multiply\\f = 20\\f = 20 cm$