Answer:
Relative to the ground, the velocity of the aircraft is 240 km/hr
Explanation:
Relative velocity is different from normal velocity;
When 2 objects are moving in opposite directions towards each other, they will appear to be faster than they actually are;
This is known as the relative velocity;
The information tells us we have the aircraft moving 320 km/hr northwards relative to the wind;
The wind is in the opposite direction at 80 km/hr;
R = relative velocity of the aircraft
v = actual velocity of the aircraft
w = velocity of the wind
R = v + w
Note: if the wind was moving in the same direction, the formula would be R = v - w
320 = v + 80
v = 320 - 80
v = 240
The velocity relative to the ground is simply the actual velocity as the ground doesn't move;
So, relative to the ground, the velocity of the aircraft is simply 240 km/hr
Answer:
See step by step sexplanation
Explanation:
1.-Sabemos que la relación:
P₁ * V₁ = P₂ * V₂
Para una temperatura constante debe mantenerse entonces si el globo se comprime hasta llevarlo a 1/3 de su valor inicial, entonces necesariamente para cumplir con la relación mencionada, la presión aumenta tres veces su valor original
2.-La definición de presión es fuerza por unidad de superficie, entonces la fuerza es determinada por la altura de la columna de liquido en el recipiente y no por la cantidad total de liquido, de acuerdo a esto habrá más presión en la base del florero, ya que la columna de agua tiene más altura.
3.-No se puede estar de acuerdo con el criterio del plomero. En su solución no plantea el aumento de la altura del tanque, para el logro del aumento de la presión que es realmente lo que hay que hacer
The formula that is applicable here is E = kQ/r^2 in which the energy of attraction is proportional to the charges and inversely proportional to the square of the distance. In this case,
kQ1/(r1)^2 = kQ2/(r2)^2 r1=l/3, r2=2l/3solve Q1/Q2
kQ1/(l/3)^2 = kQ2/(2l/3)^2 kQ1/(l^2/9) = kQ2/(4l^2/9)Q1/Q2 = 1/4
Answer:
mb = 3.75 kg
Explanation:
System of forces in balance
ΣFx =0
ΣFy = 0
Forces acting on the box
T₁ : Tension in string 1 ,at angle of 50° with the horizontal on the left
T₂ = 40 N : Tension in string 2, at angle of 75° with the horizontal on the right.
Wb :Weightt of the box (vertical downward)
x-y T₁ and T₂ components
T₁x= T₁cos50°
T₁y= T₁sin50°
T₂x= 30*cos75° = 7.76 N
T₂y= 30*sin75° = 28.98 N
Calculation of the Wb
ΣFx = 0
T₂x-T₁x = 0
T₂x=T₁x
7.76 = T₁cos50°
T₁ = 7.76 /cos50° = 12.07 N
ΣFy = 0
T₂y+T₁y-Wb = 0
28.98 + 12.07(cos50°) = Wb
Wb = 36.74 N
Calculation of the mb ( mass of the box)
Wb = mb* g
g: acceleration due to gravity = 9.8 m/s²
mb = Wb/g
mb = 36.74 /9.8
mb = 3.75 kg
