Answer:
Explanation:
acceleration of test tube
= ω² R
= (2πn)² R
= 4π²n²R
n = no of rotation per second
= 3700 / 60
= 61.67
R = .10 m
acceleration
= 4π²n²R
= 4 x 3.14² x 61.67² x .10
= 14999 N Approx
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:
d. The large pot of water and small cup of water have the same temperature, but the large pot of water has higher thermal energy.
Explanation:
Temperature is a measure of the average kinetic energy of individual molecules. While internal energy refers to the total kinetic energy of the molecules within the object. Since in this case we have the same amount of average kinetic energy, then the large pot of water and small cup of water have the same temperature. While the large pot of water has higher thermal energy, since has more water particles than the small cup.
Answer:
The frictional force 
6.446 N
The acceleration of the block a = 6.04 
Explanation:
Mass of the block = 3.9 kg
°
= 0.22
(a). The frictional force is given by
3.9 × 9.81 × 
29.3 N
Therefore the frictional force
0.22 × 29.3
6.446 N
(b). Block acceleration is given by
F = 30 N
= 6.446 N
= 30 - 6.446
= 23.554 N
The net force acting on the block is given by
23.554 = 3.9 × a
a = 6.04
This is the acceleration of the block.
The correct answer is The electromagnetic waves appear more in red color.
<span>Since red is at the low-frequency end of the visible spectrum, we say that light from a receding star is shifted toward red, or redshifted.</span>
