Answer:
F_{net-x} = 12.32 N
, F_{net-x} = -0.14 N
Explanation:
For this case we apply Newton's second law on each axis, see attachment
X axis
F cos 40 - fr = m a
Y Axis
F sin 40 +N - W = 0
the net force is
X axis
= F - fr
F_{net-x} = 20 cos 40 - 3
F_{net-x} = 12.32 N
Y Axis
F_{net-y} = F sin 40 + N - W
F_{net-x} = 20 sin 40 + 7 - 20
F_{net-x} = -0.14 N
we see that the force in this axis is approximately zero
So the air evaporate s and mix the temptures and marks air power to the same theme as the other the two degrees lower and lower until they are the same
<span>Let:
P - pushing force,
t - angle below horizontal,
F - friction force.
Resolving horizontally:
P cos(t) = F
(a)
P = F / cos(t)
= 54 / cos(30.0)
= 62.35
(b)
Work done by P
= P cos(t) * 16.0
= F * 16.0
= 54.0 * 16.0
= 864J
(c)
Work done by F
= - F * 16.0
= - 54 * 16.0
= - 864 J.
(d)
As there is no motion vertically, the work done by the gravitational force is
0.</span>
<span> </span>
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