s², what would be its final velocity. together with its associated uncertainty, ...

Questions

September 2022 1 4 Report

A. Solve the following problems.

1. The radius of a solid sphere is measured to be (6.5 ± 0.2) cm. Determine the volume of the sphere with its uncertainty.

2. A resistor is marked as having a value of 5.9 ± 2%. The power P dissipated in the resistor, when connected in a simple electrical circuit, was to be calculated from the current in the resistor, which measured as (1.40 ± 0.05) mA. What is the value of calculated P together with its associated uncertainty?

3. A car accelerates uniformly from rest and travels a distance of (100±1) m. If the acceleration of the car is (6.5 ± 0.5) m/s², what would be its final velocity. together with its associated uncertainty, at the end of the distance covered?

I will report unhelpful answers

Answers & Comments

kharmsorz

Note: I don't know what exactly your teacher have discussed on how to calculate uncertainties. But I give you two options:

(1) the simple method, and

(2) the average deviation method.

I will not show the third and fourth options which are the standard deviation method and the propagation of errors. The average deviation and standard deviation methods involve the application of statistics. The propagation of errors involves the application of calculus.

Item 1: First Option

The volume V of the sphere is

$V=\dfrac{4}{3}\pi r^3$

Assume the value of π as 3.1416.

Given:

$r_1=(6.5+0.2)\text{ cm}=6.7\text{ cm}\\r_2=(6.5-0.2)\text{ cm}=6.3\text{ cm}$

Then,

$V_1=\dfrac{4}{3}\pi r_1^3=\dfrac{4}{3}(3.1416)(6.7\text{ cm})^3\\\\\text{}\hspace{11}=1259.8\text{ cm}^3\\\\V_2=\dfrac{4}{3}\pi r_2^3=\dfrac{4}{3}(3.1416)(6.3\text{ cm})^3\\\\\text{}\hspace{11}=1047.4\text{ cm}^3$

Take the average of the volume:

$\dfrac{1259.84+1047.40}{2} =1153.6\text{ cm}^3$

Take the uncertainty:

$1259.8-1153.6=106.2\text{ cm}^3$

Thus,

$V=(1153.6\pm 106.2)\text{ cm}^3\\\boxed{V=(1200\pm 110)\text{ cm}^3}$

(in significant figures)

Item 1: Second Option

Calculating the quantity of the volume,

$V=\dfrac{4}{3} (3.1416)(6.5\;\text{cm})^3=1150.3492\;\text{cm}^3$

Calculate the uncertainty using the average deviation,

$\dfrac{\Delta V}{V}=n\dfrac{\Delta r}{r}\\\\\Delta V=Vn\dfrac{\Delta r}{r}\quad\Rightarrow n=3$

$\Delta V=(1150.3492\;\text{cm}^3)(3)\dfrac{0.2\;\text{cm}}{6.5\;\text{cm}}\\\\\Delta V=106.18608\;\text{cm}^3$

Thus,

$V\pm\Delta V\\=(1150.3492\pm 106.18606)\text{ cm}^3\\=\boxed{(1200\pm 110)\text{ cm}^3}$

(in significant figures)

Item 2: First Option

The power P with the relationship of current I and resistance R is given by

$P=I^2R$

Given:

$R_1=(5.92+0.02)\;\Omega=5.94\;\Omega\\R_2=(5.92-0.02)\;\Omega=5.90\;\Omega$

and

$I_1=(1.40+0.05)\text{ mA}=1.45\text{ mA}\\I_2=(1.40-0.05)\text{ mA}=1.35\text{ mA}$

Then,

$P_1=I_1^2R_1=(1.45\text{ mA})^2(5.94\;\Omega)\\\text{}\hspace{11.5}=12.49\;\mu\text{W}\\P_2=I_2^2R_2=(1.35\text{ mA})^2(5.90\;\Omega)\\\text{}\hspace{11.5}=10.75\;\mu\text{W}$

Take the average of the power:

$\dfrac{12.49+10.75}{2}=11.62\;\mu\text{W}$

Take the uncertainty:

$12.49-11.62=0.87\;\mu\text{W}$

Thus,

$P=(11.62\pm 0.87)\;\mu\text{W}\\\boxed{P=(12\pm 0.87)\;\mu\text{W}}$

(in significant figures)

Item 2: Second Option

Calculate the quantity of the power,

$P=I^2R=(1.40\text{ mA})^2(5.9\;\Omega)\\P=11.564\;\mu\text{W}$

Calculate the uncertainty using the average deviation,

$\dfrac{\Delta P}{P}=n\dfrac{\Delta I}{I}+\dfrac{\Delta R}{R}\\\\\Delta P=P\left(n\dfrac{\Delta I}{I}+\dfrac{\Delta R}{R}\right) \quad\Rightarrow n=2$

$\Delta P=(11.564\;\mu\text{W})\left[(2)\dfrac{0.05\;\text{mA}}{1.40\;\text{mA}}+\dfrac{0.02\;\Omega}{5.9\;\Omega}\right] \\\\\Delta P=0.8652\;\mu\text{W}$

Thus,

$P\pm\Delta P\\=(11.564\pm 0.8652)\;\mu\text{W}\\=\boxed{(12\pm 0.87)\;\mu\text{W}}$

(in significant figures)

Item 3: First Option

The kinematic equation having the acceleration a, the initial and final velocities (u and v), and the distance s is given by:

$v^2=u^2+2as$

Since the car starts at rest, the initial velocity is zero. So,

$v^2=2as\\v=\sqrt{2as}$

Given:

$s_1=(100+1.0)\;\text{m}=101\;\text{m}\\s_2=(100-1.0)\;\text{m}=99\;\text{m}$

and

$a_1=(6.5+0.5)\;\text{m/s}^2=7\;\text{m/s}^2\\a_2=(6.5-0.5)\;\text{m/s}^2=6\;\text{m/s}^2$

Then,

$v_1=\sqrt{2(7\;\text{m/s}^2)(101\;\text{m})}=37.6\;\text{m/s}\\v_2=\sqrt{2(6\;\text{m/s}^2)(99\;\text{m})}=34.5\;\text{m/s}$

Take the average of the final velocity:

$\dfrac{37.6+34.5}{2}=36.1\;\text{m/s}$

Take the uncertainty:

$37.6-36.1=1.5\;\text{m/s}$

Thus,

$v=(36.1\pm 1.5)\;\text{m/s}\\\boxed{v=(36\pm 1.5)\;\text{m/s}}$

(in significant figures)

Item 3: Second Option

Recall the formula,

$v^2=u^2+2as\quad\Rightarrow u=0\\v=\sqrt{2as}$

Calculate the quantity of the final velocity,

$v=\sqrt{2(6.5\;\text{m/s})^2(100\text{ m})}=36.05551\;\text{m/s}$

Calculate the uncertainty using the average deviation,

$\dfrac{\Delta v}{v}=m\dfrac{\Delta a}{a}+n\dfrac{\Delta s}{s}\\\\\Delta v=v\left(m\dfrac{\Delta a}{a}+n\dfrac{\Delta s}{s} \right) \quad\Rightarrow m=n=\dfrac{1}{2}$

$\Delta v=(36.05551\text{ m/s})\left[\left(\dfrac{1}{2} \right)\dfrac{0.5\;\text{m/s}^2}{6.5\;\text{m/s}^2}+\left(\dfrac{1}{2} \right)\dfrac{1\;\text{m}}{100\;\text{m}} \right]\\\\\Delta v=1.56703\;\text{m/s}$

Thus,

$v\pm\Delta v\\=(36.05551\pm 1.56703)\;\text{m/s}\\=\boxed{(36\pm 1.6)\text{ m/s}}$

(in significant figures)

kharmsorz Please refresh for updates. Thank you for your patience.

kharmsorz I got mistake in the given from Number 2.
I used 5.92 ± 2%, instead of 5.9 ± 2%

kharmsorz Can you able to delete my answer?

kharmsorz I'm going to edit the solution again.

kharmsorz Did you get a perfect score?

kharmsorz Thank goodness. Which option does your teacher refer?

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