The average radius(r) of each grain is r = 50 nanometers
= 50*10^-6 meters
Since it is spherical, so
Volume=(4/3)*pi*r^3
V= (4/3)*pi*(50*10^-6)^3
V=5.23599*10^-13 m^3
We are given the Density(ρ) =2600kg/m^3
We know that:
Density(p) = mass(m)/volume(V)
m = ρV
So the mass of a single grain is:
m = 5.23599*10^-13 * 2600 = 1.361357*10^-9 kg
The surface area of a grain is:
a = 4*pi*r^2
a = 4*pi*(50*10^-6)^2
a = 3.14*10^-8 m^2
Since we know the surface area and mass of a grain, the
conversion factor is:
1.361357*10^-9 kg / 3.14*10^-8 m^2
Find the Surface area of the cube:
cube = 6a^2
cube = 6*1.1^2 = 7.26m^2
multiply this by the converions ratio to get:
total mass of sand grains = (7.26 m^2 * 1.361357*10^-9 kg)
/ (3.14*10^-8 m^2)
total mass of sand grains = 0.3148 kg = 314.80 g
Answer:
Explanation:
Diameter of pool = 12 m
radius of pool, r = 6 m
Total height raised, h = 3 + 2.5 = 5.5 m
density of water, d = 1000 kg/m³
Mass of water, m = Volume of water x density
m = πr²h x d
m = 3.14 x 6 x 6 x 5.5 x 1000
m = 113040 kg
Work = m x g x h
W = 113040 x 9.8 x 5.5
W = 6092856 J
Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
