Winds are immigrants. They're named for where they have already been and are coming from, not for where they're going.
A Westerly wind is coming from the West. Anything caught in it is being blown toward the east.
Answer:
I would say it's B. But just in case here is some information if I'm wrong.
Explanation:
Igneous rocks are very dense and hard. They may have a glassy apprearance. Metamorphic rocks may also have a glassy appearance. You can distinguish these from igneous rocks based on the fact that metamorphic rocks tend to be brittle, lightweight, and an opaque black color.
Hope this helps!
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At surface,
v = kq/r
And potential energy of an electron is given by,
PE = -ev = -ekq/r
At escape velocity,
PE + KE = 0.
Therefore,
1/2mv^2 - ekq/r =0
1/2mv^2 = ekq/r
v = Sqrt [2ekq/mr], where v = escape velocity, e = 1.6*10^-19 C, k = 8.99*10^9 Nm^2/C^2, m = 9.11*10^-31 kg, r = 1.1*10^-2 m, q = 8*10^-9 C
Substituting;
v = Sqrt [(2*1.6*19^-19*8.99*10^9*8*10^-9)/(9.11*10^-31*1.1*10^-2)] = 47949357.23 m/s ≈ 4.795 *10^7 m/s
Answer:
C) solo III
Explanation:
Para solucionar este problema debemos analizar cada una de las opciones hasta llegar a la opcion valida.
I) el cuerpo pesa igual que su masa.
Esta opcion no puede ser ya que el peso de un cuerpo se define como el producto de la masa por la aceleracion gravitacion.
donde:
w = peso [N]
m = masa [kg]
g = aceleracion gravitacional = 9.81 [m/s²]
Como podemos ver el peso siempre sera mayar que la masa, ya que el peso es resultado de la multiplicacion de la masa por la gravedad.
II) Por medio de un analisis de fuerzas en el eje-y, la fuerza del peso se dirige hacia abajo mientras que la fuerza normal tiene igual magnitud, pero se dirige hacia arriba. Por esto la segunda opcion no puede ser.
III) El cuerpo se encuentra en equilibrio, es decir las unicas fuerzas que actuan sobre el cuerpo son el peso y la fuerza normal. Pero estas fuerzas son iguales y opuestas en direccion, por la tanto se cancelan y estan en equilibrio.
Esta es la opcion valida, la fuerza neta es nula.
Answer:
The correct option is A = 1960 N/m²
Explanation:
Given that,
Mass m= 20,000kg
Area A = 100m²
Pressure different between top and bottom
Assume the plane has reached a cruising altitude and is not changing elevation. Then sum the forces in the vertical direction is given as
∑Fy = Wp + FL = 0
where
Wp = is the weight of the plane, and
FL is the lift pushing up on the plane.
Let solve for FL since the mass of the plane is given:
Wp + FL = 0
FL = -Wp
FL = -mg
FL = -20,000× -9.81
FL = 196,200N
FL should be positive since it is opposing the weight of the plane.
Let Equate FL to the pressure differential multiplied by the area of the wings:
FL = (Pb −Pt)⋅A
where Pb and Pt are the static pressures on bottom and top of the wings, respectively
FL = ∆P • A
∆P = FL/A
∆P = 196,200 / 100
∆P = 1962 N/m²
∆P ≈ 1960 N/m²
The pressure difference between the top and bottom surface of each wing when the airplane is in flight at a constant altitude is approximately 1960 N/m². Option A is correct
