Let's break down the problem step-by-step to find (i) the pressure exerted by the block on the surface and (ii) the density of the wood.
### Given:
- Dimensions of the block: [tex]\( 20 \text{ cm} \times 7.5 \text{ cm} \times 7.5 \text{ cm} \)[/tex]
- Mass of the block: [tex]\( 1000 \text{ g} \)[/tex]
### Assumptions:
- The longest edge of the block (20 cm) is vertical.
### Steps:
#### (i) Pressure exerted by the block on the surface:
1. Convert the mass to kilograms:
The mass [tex]\( m \)[/tex] in kilograms (kg) is [tex]\( \frac{1000 \text{ g}}{1000} = 1 \text{ kg} \)[/tex].
2. Calculate the area of the base of the block:
The base area is formed by the width and height of the block. Given the dimensions, the base area [tex]\( A \)[/tex] is:
[tex]\[
\text{Base Area} = 7.5 \text{ cm} \times 7.5 \text{ cm} = 56.25 \text{ cm}^2
\][/tex]
3. Convert the area to square meters (m[tex]\(^2\)[/tex]):
Since [tex]\( 1 \text{ m}^2 = 10,000 \text{ cm}^2 \)[/tex]:
[tex]\[
\text{Base Area in} \, \text{m}^2 = \frac{56.25 \text{ cm}^2}{10,000} = 0.005625 \text{ m}^2
\][/tex]
4. Calculate the weight of the block:
Weight [tex]\( F \)[/tex] is mass [tex]\( m \)[/tex] times the acceleration due to gravity [tex]\( g \)[/tex]:
[tex]\[
F = m \times g = 1 \text{ kg} \times 9.81 \text{ m/s}^2 = 9.81 \text{ N}
\][/tex]
5. Calculate the pressure exerted on the surface:
Pressure [tex]\( P \)[/tex] is the force divided by the area:
[tex]\[
P = \frac{F}{A} = \frac{9.81 \text{ N}}{0.005625 \text{ m}^2} \approx 1744.00 \text{ Pa}
\][/tex]
The pressure exerted by the block on the surface is approximately [tex]\( 1744.00 \text{ Pascals (Pa)} \)[/tex].
#### (ii) Density of the wood:
1. Calculate the volume of the block in cubic centimeters:
Volume [tex]\( V \)[/tex] is given by:
[tex]\[
V = 20 \text{ cm} \times 7.5 \text{ cm} \times 7.5 \text{ cm} = 1125 \text{ cm}^3
\][/tex]
2. Convert the volume to cubic meters (m[tex]\(^3\)[/tex]):
Since [tex]\( 1 \text{ m}^3 = 1,000,000 \text{ cm}^3 \)[/tex]:
[tex]\[
\text{Volume in} \, \text{m}^3 = \frac{1125 \text{ cm}^3}{1,000,000} = 0.001125 \text{ m}^3
\][/tex]
3. Calculate the density of the wood:
Density [tex]\( \rho \)[/tex] is mass [tex]\( m \)[/tex] divided by volume [tex]\( V \)[/tex]:
[tex]\[
\rho = \frac{m}{V} = \frac{1 \text{ kg}}{0.001125 \text{ m}^3} \approx 888.89 \text{ kg/m}^3
\][/tex]
The density of the wood is approximately [tex]\( 888.89 \text{ kg/m}^3 \)[/tex].
### Summary:
- Pressure exerted by the block on the surface: [tex]\( 1744.00 \text{ Pa} \)[/tex]
- Density of the wood: [tex]\( 888.89 \text{ kg/m}^3 \)[/tex]
