Home Page Textbooks Physics Page 32: Lesson Check

Physics

1st Edition

Walker

ISBN: 9780133256925

Textbook solutions

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Page 32: Lesson Check

Exercise 30

Solution 1

Solution 2

Step 1

1 of 3

We define a physical quantity as $textbf{a property of an object that can be quantified by measurement.}$

Step 2

2 of 3

Three examples of a physical quantity are:

$circ$ mass

$circ$ length

$circ$ volume

Result

3 of 3

Mass, length, volume.

Step 1

1 of 2

$tt{3 examples of physical quantities are :}$

* Length
* Time
* Temperature

Result

2 of 2

$$
tt{Length,Time,Temperature}
$$

Exercise 31

Step 1

1 of 3

$textbf{When multiplying or dividing quantities with different numbers of significant figures, the rule is to assign significant figures in the result based on the smallest number of significant figures from the original set of measurements.}$

Step 2

2 of 3

If the height of the picture frame is known to three significant figures, and the width is known to two significant figures, then $textbf{the area will have two significant figures.}$

Result

3 of 3

There are two significant figures in the area of the picture frame.

Exercise 32

Step 1

1 of 2

A speed of $100 frac{text{m}}{text{s}}$ written so that it has three significant figures is

$$
begin{align*}
boxed{1.00cdot 10^{2} frac{text{m}}{text{s}}}\
end{align*}
$$

Result

2 of 2

$$
begin{align*}
boxed{1.00cdot 10^{2} frac{text{m}}{text{s}}}\
end{align*}
$$

Exercise 33

Solution 1

Solution 2

Step 1

1 of 4

The similarity between speed and velocity is that $textbf{they both describe motion and have the same unit, which, in SI, is meters per second.}$

Step 2

2 of 4

The difference between speed and velocity is that $textbf{speed is a scalar; it only has a numerical value. Velocity, however, is a vector, meaning that it has both a numerical value and a direction.}$

Step 3

3 of 4

In addition, it’s worth mentioning the connection between speed and velocity: $textbf{the magnitude (numerical value) of velocity is speed.}$

Result

4 of 4

The similarity between speed and velocity is the fact that they both describe motion. The difference, however, is that speed is a scalar, whereas velocity is a vector.

Step 1

1 of 1

The magnitude of velocity is the speed. But they are different in the sense that velocity has a direction and hence a vector while speed is just a scalar.

Exercise 34

Solution 1

Solution 2

Step 1

1 of 4

The height of the poster is $h=0.95 text{m}$, and the width is $w=1.0 text{m}$. The perimeter of the poster in question is

$$
begin{align*}
P&=2(h+w)\
&=2(0.95 text{m}+1.0 text{m})\
&=3.9 text{m}\
end{align*}
$$

Step 2

2 of 4

In this problem, both the height and the width of the poster have two significant figures, which means that $textbf{the result will also have two significant figures.}$

The number of decimals, however, is different. The height has two decimals, but the width has only one.

Step 3

3 of 4

The rule is that $textbf{the number of decimals in the final result has to be equal to the smallest number of decimal places in any of the given values.}$ Thus, the result should be written with two significant figures, and one decimal,

$$
begin{align*}
boxed{P=3.9 text{m}}\
end{align*}
$$

Result

4 of 4

The correct number of significant figures is two, and only one digit follows the decimal point;

$$
begin{align*}
boxed{P=3.9 text{m}}\
end{align*}
$$

Step 1

1 of 2

tt {The rectangle perimeter formula is $P=2h+2w$.

using the $textit{Rule for Addition and Subtraction}$ we deduce that result must have 2 Decimal spaces.

Result

2 of 2

$$
tt{ the result has 2 Decimal spaces}
$$

Exercise 35

Step 1

1 of 3

The speed of light to 5 significant figures is $c=2.9979cdot 10^{8} frac{text{m}}{text{s}}.$ If we want to write the speed of light with 3 significant figures, the result will have two decimals, but it isn’t just simply $c=2.99cdot 10^{8} frac{text{m}}{text{s}}$. Instead, we have some rounding off to do.

Step 2

2 of 3

The first digit to be dropped (in this case 7) is greater than or equal to 5, which means that $textbf{the previous digit (the second 9) has to be increased by 1}$ (for 9, we write 0). In addition to that, $textbf{we also have to increase the first digit (also 9) by 1, as well as the digit preceding the decimal point, which is 2.}$ Thus, the correct way to wright the speed of light with 3 significant figures is

$$
begin{align*}
boxed{c=3.00cdot 10^{8} frac{text{m}}{text{s}}}\
end{align*}
$$

Result

3 of 3

$$
begin{align*}
boxed{c=3.00cdot 10^{8} frac{text{m}}{text{s}}}\
end{align*}
$$

Exercise 36

Solution 1

Solution 2

Step 1

1 of 3

The initial speed of the bus is $v_{i}=2.2 frac{text{m}}{text{s}}$, and after $20 text{s}$ the speed increases by $5.225 frac{text{m}}{text{s}}$. Our task is to find the final speed of the bus, $v_{f}$.

Step 2

2 of 3

The correct result for this problem should have $textbf{two significant figures,}$ corresponding to the quantity with the smallest number of significant figures ($v_{i}$), and only $textbf{one digit after the decimal point,}$ corresponding to the input value with the smallest number of decimals (also $v_{i}$). Therefore, the speed of the bus after 20 seconds is

$$
begin{align*}
v_{f}&=2.2 dfrac{text{m}}{text{s}}+5.225 dfrac{text{m}}{text{s}}\
&=quadboxed{7.4 dfrac{text{m}}{text{s}}}\
end{align*}
$$

Result

3 of 3

$$
begin{align*}
boxed{v_{f}=7.4 dfrac{text{m}}{text{s}}}\
end{align*}
$$

Step 1

1 of 2

Initial speed of bus $=(2.2)m/s$

Increase in speed of bus $=(5.225)m/s$

So, after $20s$ , final speed of bus $=(2.2+5.225)m/s=(7.425)m/s=boxed{(7.4)m/s}$

The final result has only one digit after decimal point, corresponding to value with least number of significant digits , i.e. $(2.2)m/s$ .

Result

2 of 2

$(7.4)$ $m/s$

Exercise 37

Solution 1

Solution 2

Step 1

1 of 4

$textbf{(a)}$ The area of the screen is given by $heighttimes width$, the height and width being $h=31.25 text{cm}$ and $w=47 text{cm}$, respectively. $textbf{Following the rule for multiplication of two values that differ in the number of significant figures, the area is known to two significant figures, because the least accurately known quantity has two significant figures.}$

Step 2

2 of 4

$textbf{(b)}$ Calculating the area of the screen, we get the following result:

$$
begin{align*}
A&=htimes w\
&=31.25 text{cm}times 47 text{cm}\
&=1468.75 text{cm}^{2}\
end{align*}
$$

This, however, is neither the correct number of significant figures, nor the correct number of decimals. $textbf{The correct answer should have two significant digits and no decimals}$ (because the least accurate quantity has no decimals). Therefore, the area with the correct number of significant figures is

$$
begin{align*}
boxed{A=1500 text{cm}^{2}}\
end{align*}
$$

Step 3

3 of 4

* Notice that we rounded off the result in $textbf{(b)}$ to $1500 text{cm}^{2}$, rather then $1400 text{cm}^{2}$, the reason being that the rough estimate of $1468 text{cm}^{2}$ is $1500 text{cm}^{2}$, since 1468 is greater than 1450. If the area were something like $1423 text{cm}^{2}$, we would round it down to $1400 text{cm}^{2}$.

Result

4 of 4

$textbf{(a)}$ The area of the screen is known to two significant figures.

$textbf{(b)}$ $boxed{A=1500 text{cm}^{2}}$

Step 1

1 of 2

$a)quad$As per rule for multiplication, the number of significant figures in the answer is the same as the number of significant figures in the least accurately known input value.

Now, for the screen :

Hight $(h)=(31.25)cm$

Width $(w)=(47)cm$

$(47)cm$ has the least , i.e. $2$ significant digits

Area $=(htimes w)quad$will have $2$ significant digits

$b)$

Area of screen, $A=(htimes w)=(31.25times47)cm^2=(1468.75)cm^2$

In correct significant digits (i.e. $2$) , area , $A=boxed{1500cm^2}$

Result

2 of 2

$a)quad 2$

$b)quad1500$ $cm^2$

Exercise 38

Solution 1

Solution 2

Step 1

1 of 4

$textbf{(a)}$ The perimeter of the parking lot, with the correct number of significant figures and decimals, is

$$
begin{align*}
P&=2(l+ w)\
&=2(144.3 text{m}+47.66 text{m})\
&=quadboxed{383.9 text{m}}\
end{align*}
$$

where $l$ is the length of the lot, and $w$ is the width.

Step 2

2 of 4

$textbf{(b)}$ The area of the parking lot, with the correct number of significant figures, is

$$
begin{align*}
A&=ltimes w\
&=144.3 text{m}times 47.66 text{m}\
&=quadboxed{6877 text{m}^{2}}\
end{align*}
$$

with, again, $l$ being the length, and $w$ being the width of the lot.

Step 3

3 of 4

* Notice that there are no decimals in the solution for $textbf{(b)}$, because the rule for the number of significant figures outweighs the rule for the number of decimals.

Result

4 of 4

$textbf{(a)}$ $boxed{383.9 text{m}}$

$textbf{(b)}$ $boxed{P=6877 text{m}^{2}}$

Step 1

1 of 2

Parking lot : length $=(144.3)m=L(text{say})$

. $qquadqquad$ width $=(47.66)m=omega(text{say})$

$a)quad$As per rule for addition, perimeter $=2(L+omega)=2(144.3+47.66)m$

.$qquadqquadqquadqquadqquadqquadqquadqquadquad=(383.92)m=boxed{(383.9)m}$

$b)quad$As per rule for multiplication, area $=(Ltimesomega)=(144.3times47.66)m^2$

.$qquadqquadqquadqquadqquadqquadqquadqquadquad=(6877.338)m^2=boxed{(6877)m^2}$

Result

2 of 2

$a)quad383.9$ $m$

$b)quad 6877$ $m^2$

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Page 32: Lesson Check

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