Answer:
If it isn't an ion it should have 35 elektrons to cancel the positivity of the nucleus.
Diagram a shows you a plate tectonic not a boundary and diagram b shows you a convergent plate boundary because the plates are coming together and the oceanic plate is being subducted undneath the continental plate
The closer the particles, the more will be the propogation of sound waves. Room contains air molecules which are far away from each other. So it takes much time for one molecule of air to disturb the other one. But in case of solids, as particles are much closer(compared to fluids), disturbance generated by one molecule is quickly transmitted to the next molecule
Answer:
v₂ = 63.62 m / s
Explanation:
For this exercise in fluid mechanics we will use Bernoulli's equation
P₁ + ρ g v₁² + ρ g y₁ = P₂ + ρ g v₂² + ρ g y₂
where the subscript 1 refers to the inside of the wing and the subscript 2 to the top of the wing.
We will assume that the distance between the two parts is small, so y₁ = y₂
P₁-P₂ = ρ g (v₂² - v₁²)
pressure is defined by
P = F / A
we substitute
ΔF / A = ρ g (v₂² - v₁²)
v₂² = 
suppose that the area of the wing is A = 1 m²
we substitute
v₂² = 
v₂² = 79.10 + 3969
v₂ = √4048.1
v₂ = 63.62 m / s
Answer:
t1 = t2 + 3.02 V = 41.5
V t1 - 1/2 g t1^2 = V t2 - 1/2 g t2^2
Both stones reach the same height after the specified times
V (t1 - t2) = g/2 (t1^2 - t2^2) = g/2 (t1 - t2) (t1 + t2)
2 V / g = t1 + t2 = 2t1 + 3.02
t1 = V / g - 1.51 = 41.5 / 9.8 -1.51 = 2.72 s
t2 = t1 + 3.02 = 5.74 sec
Check:
41.5 * 2.72 - 4.9 * 2.72^2 = 76.6 m
41.5 * 5.74 - 4.9 * 5.74^2 = 76.8 m
Speed of second stone = 41.5 - 9.8 * 2.72 = 14.8 m/s
