Answer:
During the first quarter or last quarter phase of the moon, when the sun and moon are perpendicular (at right angles) to each other in relation to the Earth, the tidal gravitational pulls interfere with each other, producing weaker tides, known as neap tides.
Explanation:
Answer: The average velocity is 150 km/h
Explanation: 70+80=150
Answer:
λ = 538.0 nm
Explanation:
The solution of the Schrödinger equation for the inner part of the well gives energy
= (h² / 8mL²) n²
Where n is an integer and L is the length of the well
They ask for the transition from the first excited state n = 2 to the base state n = 1
E₂ - E₁ = = (h² / 8mL²) (n₂² - n₁²)
Let's calculate
E₂-E₁ = (6.63 10⁻³⁴)² / (8 9.1 10⁻³¹ (0.7 10⁻⁹)²) (2² -1²)
E₂ –E₁ = 3.6968 10⁻¹⁹ J
Let's use the Planck equation
E = h f
c = λ f
E = h c / λ
E = E₂ ₂- E₁
h c / λ = 3.6968 10⁻¹⁹
λ = h c / (E₂-E₁)
λ = 6.63 10⁻³⁴ 3 10⁸ / 3.6968 10⁻¹⁹
λ = 5.380 10⁻⁷ m
Let's reduce
λ = 5.380 10⁻⁷ m (10 9 nm / 1 m)
λ = 538.0 nm
Answer:
a) 
b) 
Explanation:
Given:
height of water in one arm of the u-tube, 
a)
Gauge pressure at the water-mercury interface,:

we've the density of the water 


b)
Now the same pressure is balanced by the mercury column in the other arm of the tube:



<u>Now the difference in the column is :</u>



Answer:
V = 1060.8 cm³
Explanation:
we know that the pressure,
P = density x gravity x depth
ρm is the density of mercury
ρw is the density of water
the pressure due to mercury P₁= (ρm) g h₁
the pressure due to water P₂ = (ρw) g h₂
the total pressure
P = P₁ + P₂
P = (ρm) g h₁ + (ρw) g h₂
but the total pressure is double the pressure due to mercury.
(ρm) g h₁ =(ρw) g h₂


h₂ = 81.6 cm
the height of the water is 81.6 cm
the volume
V = height x area
V = 81.6 x 13
V = 1060.8 cm³
the volume of water must be added to double the gauge pressure is V = 1060.8 cm³