Yea it would be barrier. since the stream has a cover for the water
Answer: the contents of this container weighs 4905 kg.m/s²
Explanation:
Given that;
volume of a container V = 0.5 m³
we know that standard gravitational acceleration g = 9.81 m/s²
specific volume of liquid filled in the container v = 0.001 m³/kg
now we express the equation for weight of the container.
W = mg
W = (pV)g
W = Vg / ν
so we substitute
W = (0.5 m³)(9.81 m/s ) / 0.001 m³/kg
W = 4.905 / 0.001
W = 4905 kg.m/s²
Therefore, the contents of this container weighs 4905 kg.m/s²
Answer:
1.98 x 10⁻⁷ F
Explanation:
w = width of the sheet = 5.9 cm = 0.059 m
L = length of the sheet = 5.6 m
Area of the sheet is given as
A = L w = (5.6) (0.059) = 0.3304 m²
d = distance between the sheets = 3.1 x 10⁻⁵ m
k = dielectric constant of teflon = 2.1
Capacitance is given as


C = 1.98 x 10⁻⁷ F
C. The water cycle spreads water out evenly around the whole Earth
Answer:
47 m
Explanation:
Data obtained from the question include the following:
Length of dry leg 1 (L1) = 40 m
Length of dry leg 2 (L2) = 25 m
Length of swimming course (L) =..?
The length of the swimming course can be obtained by using pythagoras theory as shown below:
L² = L1² + L2²
L² = 40² + 25²
L² = 1600 + 625
L² = 2225
Take the square root of both side.
L = √2225
L = 47.1 ≈ 47 m
Therefore, the length of the swimming course is approximately 47 m.