25. the sides of a small rectangular box are measured to be 1.80 ± 0.01 cm, 2.05 ± 0.02 cm, and 3.1 ± 0.1 cm…

25. the sides of a small rectangular box are measured to be 1.80 ± 0.01 cm, 2.05 ± 0.02 cm, and 3.1 ± 0.1 cm long. calculate its volume and uncertainty in cubic centimeters.\n26. when non - metric units were used in the united kingdom, a unit of mass called the pound - mass (lbm) was employed, where 1 lbm = 0.4539 kg. (a) if there is an uncertainty of 0.0001 kg in the pound - mass unit, what is its percent uncertainty? (b) based on that percent uncertainty, what mass in pound - mass has an uncertainty of 1 kg when converted to kilograms?\n27. the length and width of a rectangular room are measured to be 3.955 ± 0.005 m and 3.050 ± 0.005 m. calculate the area of the room and its uncertainty in square meters.

25. the sides of a small rectangular box are measured to be 1.80 ± 0.01 cm, 2.05 ± 0.02 cm, and 3.1 ± 0.1 cm long. calculate its volume and uncertainty in cubic centimeters.\n26. when non - metric units were used in the united kingdom, a unit of mass called the pound - mass (lbm) was employed, where 1 lbm = 0.4539 kg. (a) if there is an uncertainty of 0.0001 kg in the pound - mass unit, what is its percent uncertainty? (b) based on that percent uncertainty, what mass in pound - mass has an uncertainty of 1 kg when converted to kilograms?\n27. the length and width of a rectangular room are measured to be 3.955 ± 0.005 m and 3.050 ± 0.005 m. calculate the area of the room and its uncertainty in square meters.

Answer

25.

Explanation:

Step1: Calculate the volume

The volume $V$ of a rectangular box with sides $a = 1.80$ cm, $b=2.05$ cm and $c = 3.1$ cm is given by $V=a\times b\times c$. $V=1.80\times2.05\times3.1=11.391$ $cm^{3}$

Step2: Calculate the relative uncertainties

The relative uncertainty in a product $y = x_1\times x_2\times\cdots\times x_n$ is $\frac{\Delta y}{y}=\sqrt{(\frac{\Delta x_1}{x_1})^2+(\frac{\Delta x_2}{x_2})^2+\cdots+(\frac{\Delta x_n}{x_n})^2}$ For $a = 1.80\pm0.01$ cm, $\frac{\Delta a}{a}=\frac{0.01}{1.80}$; for $b = 2.05\pm0.02$ cm, $\frac{\Delta b}{b}=\frac{0.02}{2.05}$; for $c=3.1\pm0.1$ cm, $\frac{\Delta c}{c}=\frac{0.1}{3.1}$ $\frac{\Delta a}{a}=\frac{0.01}{1.80}\approx0.00556$, $\frac{\Delta b}{b}=\frac{0.02}{2.05}\approx0.00976$, $\frac{\Delta c}{c}=\frac{0.1}{3.1}\approx0.0323$ $\frac{\Delta V}{V}=\sqrt{(0.00556)^2+(0.00976)^2+(0.0323)^2}$ $=\sqrt{0.0000309+0.0000953+0.0010433}=\sqrt{0.0011695}\approx0.0342$

Step3: Calculate the uncertainty in volume

$\Delta V = V\times\frac{\Delta V}{V}=11.391\times0.0342\approx0.39$ $cm^{3}$

Answer:

The volume is $V = 11.4\pm0.4$ $cm^{3}$

26.

(a)

Explanation:

Step1: Calculate the percent - uncertainty

The percent - uncertainty is given by $\text{Percent uncertainty}=\frac{\Delta m}{m}\times100%$ Given $m = 0.4539$ kg and $\Delta m=0.0001$ kg $\text{Percent uncertainty}=\frac{0.0001}{0.4539}\times100%\approx0.022%$

Answer:

The percent uncertainty is $0.022%$

(b)

Explanation:

Step1: Let the mass in pound - mass be $m_{lbm}$

We know that the percent uncertainty is constant. Let the mass in kg be $m_{kg}$ and its uncertainty $\Delta m_{kg}=1$ kg. Since $\frac{\Delta m_{kg}}{m_{kg}}=\frac{\Delta m_{lbm}}{m_{lbm}}$ and we know the percent uncertainty from part (a) is $\frac{\Delta m_{lbm}}{m_{lbm}}=\frac{0.0001}{0.4539}$ If $\Delta m_{kg} = 1$ kg, then $m_{kg}=\frac{m_{lbm}\times\Delta m_{kg}}{\Delta m_{lbm}}$ Since $\frac{\Delta m_{lbm}}{m_{lbm}}=\frac{0.0001}{0.4539}$, we can rewrite it as $m_{lbm}=\frac{m_{kg}\times\Delta m_{lbm}}{\Delta m_{kg}}$ We know that $1$ lbm $ = 0.4539$ kg. Let the mass in lbm be $x$. Its mass in kg is $0.4539x$. The percent uncertainty is the same, so $\frac{1}{0.4539x}=\frac{0.0001}{0.4539}$ $x=\frac{1}{0.0001}=10000$ lbm

Answer:

The mass in pound - mass is $10000$ lbm

27.

Explanation:

Step1: Calculate the area

The area $A$ of a rectangle with length $l = 3.955$ m and width $w = 3.050$ m is $A=l\times w$ $A=3.955\times3.050 = 12.06275$ $m^{2}$

Step2: Calculate the relative uncertainties

The relative uncertainty in a product $y=x_1\times x_2$ is $\frac{\Delta y}{y}=\sqrt{(\frac{\Delta x_1}{x_1})^2+(\frac{\Delta x_2}{x_2})^2}$ For $l = 3.955\pm0.005$ m, $\frac{\Delta l}{l}=\frac{0.005}{3.955}$; for $w = 3.050\pm0.005$ m, $\frac{\Delta w}{w}=\frac{0.005}{3.050}$ $\frac{\Delta l}{l}=\frac{0.005}{3.955}\approx0.00126$, $\frac{\Delta w}{w}=\frac{0.005}{3.050}\approx0.00164$ $\frac{\Delta A}{A}=\sqrt{(0.00126)^2+(0.00164)^2}=\sqrt{0.00000159+0.00000269}=\sqrt{0.00000428}\approx0.00207$

Step3: Calculate the uncertainty in area

$\Delta A=A\times\frac{\Delta A}{A}=12.06275\times0.00207\approx0.025$ $m^{2}$

Answer:

The area is $A = 12.06\pm0.03$ $m^{2}$