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
I believe Box B will have a greater gravitational pull because the gravitational pull of an object depends on its mass. The more mass an object has, the greater its gravitational pull will become.
For example, we can take planets. Naturally, they are round because once upon a time there was a larger piece of rock that attracted others. But the size of the rock won't matter, it's the weight that matters. If the rock weighed nothing, the other rocks would just rebound upon contact. But if the rock weighed a lot, then things wouldn't so easily rebound and might actually stick to it.
It’s D. An enlargement (hope this helps!)
Answer:
Explanation:
This is an application of Newton's second Law.
Formula
F = m * a
F = 300 N
m = 100 kg
a = ?
F = m * a
300N = 100 kg * a Divide by 100
300N/100kg = a
a = 3 m/sec^2
The answer is A
Good luck!
