4.
Medial collateral ligament, which runs along inside of the knee.
Lateral collateral ligament, which runs along the outside of the knee.
Ligament, which runs in the middle part of the knee.
Posterior cruciate ligament, which works together with the Anterior cruciate ligament.
Answer:
15.0 µm
Step-by-step explanation:
Density = mass/volume
D = m/V Multiply each side by V
DV = m Divide each side by D
V = m/D
Data:
m = 1.091 g
D = 7.28 g/cm³
l = 10.0 cm
w = 10.0 cm
Calculation:
<em>(a) Volume of foil
</em>
V = 1.091 g × (1 cm³/7.28 g)
= 0.1499 cm³
(b) <em>Thickness of foil
</em>
The foil is a rectangular solid.
V = lwh Divide each side by lw
h = V/(lw)
= 0.1499/(10 × 10)
= 1.50 × 10⁻³ cm Convert to millimetres
= 0.015 mm Convert to micrometres
= 15.0 µm
The foil is 15.0 µm thick.
Polyethene is a polymer composed of repeating units of the monomer ethene.
The properties of polyethene are as follows:
- density- ranges 0.857 g/cm3 to 0.975 g/cm3.
- specific heat capacity is 1.9 kJ/kg.
- melting temperature is approximately 110 °C.
<h3>What are polymers?</h3>
Polymers are large macromolecules consisting of long repeating chains of smaller molecules known as monomers.
An example of a polymer is polyethene composed of repeating units of the monomer ethene.
The density of polyethylene ranges 0.857 g/cm3 to 0.975 g/cm3.
The specific heat capacity of polyethene is 1.9 kJ/kg.
The melting temperature of polyethene is approximately 110 °C.
Learn more about polyethene at: brainly.com/question/165779
Answer:
252.68 K or -20.46 °C
Explanation:
According to Gay-Lussac's Law, "Pressure and Temperature at given volume are directly proportional to each other".
Mathematically,
P₁ / T₁ = P₂ / T₂ ---- (1)
Data Given:
P₁ = 30.7 kPa
T₁ = 0.00 °C = 273.15 K
P₂ = 28.4 kPa
T₂ = <u>???</u>
Solving equation for T₂,
T₂ = P₂ T₁ / P₁
Putting values,
T₂ = 28.4 kPa × 273.15 K / 30.7 kPa
T₂ = 252.68 K or -20.46 °C
