Answer- thermal energy :)
B is the correct answer according to my calculations.
Answer:
Weight required = 194.51 N
Explanation:
The elongation is given by
Length , L= 1.6 m
Diameter, d = 1.1 mm
Area
Change in length, ΔL = 2.8 mm = 0.0028 m
Young's modulus of copper, E = 117 GPa = 117 x 10⁹ Pa
Substituting,
Weight required = 194.51 N
Answer:
19320 kg/m³
Explanation:
density: This can be defined as the ratio of the mass of a body to its volume. The S.I unit of Density is kg/m³.
The formula of density is given as,
D = m/v ......................... Equation 1.
Where D = Density of the gold, m = mass of the gold, v = volume of the gold.
Note: From Archimedes's Principle, the piece of gold displace an amount of water that is equal to it's volume.
Amount of water displace = 27.2 - 25 = 2.2 mL.
Given: m = 42.504 g = 0.042504 kg, v = 2.2 mL = (2.2/10⁶) m³ = 0.0000022 m³
Substitute into equation 1
D = 0.042504/0.0000022
D = 19320 kg/m³
Hence the density of the piece of gold = 19320 kg/m³
The pressure at the bottom of the tank is given by:
P = ρgh
P = pressure, ρ = fluid density, g = gravitational acceleration, h = depth
Given values:
ρ = 820kg/m³, g = 9.8m/s², h = 4.5m
Plug in and solve for P:
P = 820(9.8)(4.5)
P = 36000Pa
