Home Page Textbooks Physics Page 41: Standardized Test Prep

Physics

1st Edition

Walker

ISBN: 9780133256925

Textbook solutions

All Solutions

Page 41: Standardized Test Prep

Exercise 1

Step 1

1 of 2

The equation for the kinetic energy is given by $KE=frac{1}{2}mv^{2}$. Solving for $v$ yields

$$
begin{align*}
KE&=dfrac{1}{2}mv^{2}\
v^{2}&=left. dfrac{2(KE)}{m}quadquadright/ sqrt{quad}\
rightarrow &quadboxed{v=sqrt{dfrac{2(KE)}{m}}}\
end{align*}
$$

Therefore, the correct answer is $textbf{(B)}$.

Result

2 of 2

$textbf{(B)}$

Exercise 2

Step 1

1 of 2

The SI base unit of time is the second, thus the correct answer is $textbf{(B)}$.

Result

2 of 2

$textbf{(B)}$

Exercise 3

Solution 1

Solution 2

Step 1

1 of 2

The SI base units of measurement in this question are kilogram, second and meter. The unit that isn’t an SI base unit is kilometer, therefore the answer is $textbf{(C)}$.

Result

2 of 2

$textbf{(C)}$

Step 1

1 of 2

Kilometer is not the SI unit of length. It is meter.

Result

2 of 2

(C)

Exercise 4

Solution 1

Solution 2

Step 1

1 of 6

A detailed dimensional analysis is written below for each option in question.

Step 2

2 of 6

$textbf{(A)}$ $d=vt+frac{1}{2}at^{2}$

$left[text{L}right]=left[dfrac{text{L}}{text{T}}right]left[text{T}right] +left[dfrac{text{L}}{text{T}^{2}}right]left[text{T}^{2}right]$

$left[text{L}right]=left[text{L}right]+left[text{L}right]$

Step 3

3 of 6

$textbf{(B)}$ $v=at$

$left[dfrac{text{L}}{text{T}}right]=left[dfrac{text{L}}{text{T}^{2}}right]left[text{T}right]$

$left[dfrac{text{L}}{text{T}}right]=left[dfrac{text{L}}{text{T}}right]$

Step 4

4 of 6

$textbf{(C)}$ $v^{2}=frac{1}{2}ad^{2}$

$left[dfrac{text{L}^{2}}{text{T}^{2}}right]=left[dfrac{text{L}}{text{T}^{2}}right]left[text{L}^{2}right]$

$left[dfrac{text{L}^{2}}{text{T}^{2}}right]neqleft[dfrac{text{L}^{3}}{text{T}^{2}}right]$

Step 5

5 of 6

$textbf{(D)}$ $v^{2}=2ad$

$left[dfrac{text{L}^{2}}{text{T}^{2}}right]=left[dfrac{text{L}}{text{T}^{2}}right]left[text{L}right]$

$left[dfrac{text{L}^{2}}{text{T}^{2}}right]=left[dfrac{text{L}^{2}}{text{T}^{2}}right]$

Result

6 of 6

$textbf{(C)}$

Step 1

1 of 2

In option (C), LHS has dimensions of $v^2 = (length/time)^2$.

RHS has dimensions $(1/2)ad^2 = (length)^3/(time)^2$.

They do not match.

Result

2 of 2

(C)

Exercise 5

Solution 1

Solution 2

Step 1

1 of 2

The answer is $textbf{(D)}$. We know that $1 text{m}=100 text{cm}$. Furthermore, the prefix $textit{centi}$ means one hundredth, or $10^{-2}$, and 1 centimeter is one hundredth of a meter. Therefore, $150 text{cm}$ is then $1.5 text{m}$.

Result

2 of 2

$textbf{(D)}$

Step 1

1 of 2

In option (D), centimeter has the SI unit prefix centi- and is correctly related to the SI unit meter because 1 meter = 100 centimeter.

Result

2 of 2

(D)

Exercise 6

Step 1

1 of 2

Case by case, we have:

$textbf{(A)}$ $1.00cdot 10^{4} text{cm}=10 000 text{cm}=100 text{m}$

$textbf{(B)}$ $1.00cdot 10^{-1} text{km}=0.1 text{km}=100 text{m}$

$textbf{(C)}$ $1.00cdot 10^{2} text{m}=100 text{m}$

$textbf{(D)}$ $1.00cdot 10^{6} text{mm}=1 000 000 text{mm}=1000 text{m}neq 100 text{m}$

Therefore, the answer is $textbf{(D)}$.

Result

2 of 2

$textbf{(D)}$

Exercise 7

Solution 1

Solution 2

Step 1

1 of 2

The reasonable estimate would be the option $textbf{(B)}$. 1000 golf balls is most probably not enough to fill a home swimming pool, and $10^{8}$ is already too much, which leaves $10^{5}$.

Result

2 of 2

$textbf{(B)}$

Step 1

1 of 2

1000 balls are less for a swimming pool. For a standard size swimming pool, option (B) sounds most reasonable though on a slightly larger scale.

Result

2 of 2

(B)

Exercise 8

Step 1

1 of 2

This graph is definitely not linear, so we can eliminate options $textbf{(B)}$ and $textbf{(C)}$. The easiest way to see which of the remaining options is correct is to take $x=10$ and see to which value of $y$ it corresponds.

So, for $x=10$, $y=200$. That means that the equation $textbf{(A)}$ describes the line on the graph.

Result

2 of 2

$textbf{(A)}$

Exercise 9

Solution 1

Solution 2

Step 1

1 of 2

Just by looking at the graph, the estimated value for $y$ when $x=3$ would be somewhere around 20. This is easily checked by plugging $x=3$ into the equation that describes the line on the graph, which is $y=2x^{2}$.

All that being said, the correct answer is $textbf{(C)}$.

Result

2 of 2

$textbf{(C)}$

Step 1

1 of 2

It looks close to twenty and it can be verified by plugging x=3 in $y=2x^2$ to find actual value is 18.

Result

2 of 2

(C)

Exercise 10

Solution 1

Solution 2

Step 1

1 of 3

$textbf{(a)}$ To find the average length, simply take the sum of all recorded lengths, and divide it by the number of measurements. The correct number of significant figures in the result should correspond to the number of significant figures of the least accurately known measurement, which is two.

$$
begin{align*}
l_{avg}&=dfrac{l_{1}+l_{2}+l_{3}+l_{4}+l_{5}}{5}\
&=dfrac{12.2 text{cm}+12.1 text{cm}+12 text{cm}+11.9 text{cm}+12.20 text{cm}}{5}\
&=quadboxed{12 text{cm}}\
end{align*}
$$

Step 2

2 of 3

$textbf{(b)}$ In general, it’s best to have as much significant figures as possible with your measuring equipment. This leads to overall more accurate results of your measurements.

Based on the result in part $textbf{(a)}$, you can conclude the same thing. Different measurements have different numbers of significant figures, and your final result can’t be more accurate than the least accurate input value, i.e., if all measurements had four significant figures, the average calculated length would differ from the one we got, and would be a more precise representation of the actual length.

Result

3 of 3

$textbf{(a)}$ $boxed{l_{avg}=12 text{cm}}$

$textbf{(b)}$ Recording all measurements to the appropriate number of significant figures leads to a more precise result.

Step 1

1 of 2

$$
tt{(a)the average length is: }
$$

$$
frac{12.2+12.1+12+11.9+12.20}{5}=12.08cm
$$

using the $textit{Rule of addition and substraction}$ on the nominator we only have two significant digits

The result should be then: $12cm$

(b) writing the result with the right number of significant digits allow refers to the accuracy and precision of the measurements overall.

Result

2 of 2

$$
tt{(a) 12cm,(b) Accuracy and precision notions. }
$$

unlock

All Solutions

Page 41: Standardized Test Prep

Haven't found what you were looking for?

Search for samples, answers to your questions and flashcards