Answer:
400 m
Explanation:
The swimmer swims 50 meters to one end of the pool but has to swim back therefore you double 50 which would be 100 meters. Then you have to multiply 100 by 4 since the swimmer did it 4 times.
Answer:
The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules
Explanation:
The given spring constant of the of the spring, k = 88.0 N/m
The length by which the hose is stretched, x = 4.20 m
For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose
The elastic potential energy, P.E., of a compressed spring is given as follows;
P.E. = 1/2·k·x²
∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²
1/2 × 88.0 N/m × (4.20 m)² = 776.16 J
The work done on the hose = The potential energy given to hose, P.E. = 776.16 J
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
In 1920, after returning from Army service, he produced a successful model and in 1923 turned it over to the Northeast Electric Company of Rochester for development.
Answer:
It is Conductivity because it is the measure of the ease.
