<em>HERE'S YOUR ANSWER...</em>
• In coastal areas have less variation in temperature that. non-coastal area due to sea-breeze and land-breeze.
<em>EXPLANATION:</em>
• In day time the land heats up faster than the sea, so air above sand gets heated and rises up and cool air above ocean moves towards land to take its place.
• At night the land cools down faster than sea which causes hot air above sea to rise up and cool air above land moves in to take its place.
<em>HOPE</em><em> </em><em>IT</em><em> </em><em>HELPS</em><em>.</em><em>.</em><em>.</em>
The following apply:
Sound is a LONGITUDINAL wave.
Sound is a PRESSURE wave.
Longitudinal waves are waves in which the displacement of the medium is in the same direction as or opposite direction to the direction of wave propagation. An example of longitudinal wave is sound wave. Sound wave is also considered as a pressure wave as the result of the presence of repeating patterns of high and low pressure regions moving through a medium.
Answer:
a) 0.15 μC b) 9.4*10¹¹ electrons.
Explanation:
As the total charge must be conserved, the total charge on the spheres, after being brought to contact each other, and then separated, must be equal to the total charge present in the spheres prior to be put in contact:
Q = +8.2μC +9.0 μC +(-7.8 μC) + (-8.8 μC) = +0.6 μC
As the spheres are assumed perfect conductors, as they are identical, once in contact each other, the excess charge spreads evenly on each sphere, so the final charge, on each of them, is just the fourth part of the total charge:
Qs = Qt/4 = 0.6 μC / 4 = 0.15 μC.
b) As the charge has a positive sign, this means that each sphere has a defect of electrons.
In order to know how many electrons are absent in each sphere, we can divide the total charge by the charge of one electron, which is the elementary charge e, as follows:
Your answer is.......... That's because Earth is a solid mass and the Sun is gaseous.
Answer:
P₂ = 392720.38 Pa = 392.72 kPa
Explanation:
Given
D₁ = 5 cm = 0.05 m
D₂ = 10 cm = 0.10 m
v₁ = 8 m/s
P₁ = 380 kPa = 380000 Pa
α = 1.06
ρ = 1000 kg/m³
g = 9.8 m/s²
We can use the following formula
(P₁ / (ρg)) + α*(V₁² / (2g)) + z₁ = (P₂ / (ρg)) + α*(V₂² / (2g)) + z₂ + +hL
knowing that z₁ = z₂ we have
(P₁ / (ρg)) + α*(V₁² / (2g)) = (P₂ / (ρg)) + α*(V₂² / (2g)) + +hL <em> (I)
</em>
Where
V₂ can be obtained as follows
V₁*A₁ = V₂*A₂ ⇒ V₁*( π* D₁² / 4) = V₂*( π* D₂² / 4)
⇒ V₂ = V₁*(D₁² / D₂²) = (8 m/s)* ((0.05 m)² / (0.10 m)²)
⇒ V₂ = 2 m/s
and
hL is a head loss factor: hL = α*(1 - (D₁² / D₂²))²*v₁² / (2*g)
⇒ hL = (1.06)*(1 – ((0.05 m) ² / (0.10 m)²))²*(8 m/s)² / (2*9.8 m/s²)
⇒ hL = 1.9469 m
Finally we get P₂ using the equation <em>(I)
</em>
⇒ P₂ = P₁ - ((V₂² - V₁²)* α*ρ / 2) – (ρ*g* hL)
⇒ P₂ = 380000 Pa - (((2 m/s)² - (8 m/s)²)*(1.06)*(1000 kg/m³) / 2) – (1000 kg/m³*9.8 m/s²*1.9469 m)
⇒ P₂ = 392720.38 Pa = 392.72 kPa