Answer:
a) correct answer is C
, b) 14º from the west to the north, c) v_{1g} = 300.79 km / h
Explanation:
This is a relative speed exercise using the addition of speeds.
1) when it is not specified regarding what is being measured, the medicine is carried out with respect to the Z Earth, therefore the correct answer is C
2 and 3) In this case we must compose the speed using the Pythagorean Theorem.
² = 
² + 
²
where v_{1a} is the speed of the airplane with respect to the air, v_{1g} airplane speed with respect to the Earth, v_{ag} air speed with respect to the Earth
in this case let's clear the speed of the airplane with respect to the Earth
v_{1g} = √(v_{1a}² - v_{ag}²)
v_{1g} = √ (310² - 75²)
v_{1g} = 300.79 km / h
we find the direction of the airplane using trigonometry
sin θ = v_{ag} / v_{1a}
θ = sin⁻¹ (v_{ag} /v_{1a})
θ = sin⁻¹ (75/310)
θ= 14º
the pilot must direct the aircraft at an angle of 14º from the west to the north
What fraction of stipend triangle is a shaded triangle? What fraction of the spotted triangle is a shaded triangle? Use >,
Answer:Velocity = 6.325m/s
Directional angle= 18.43°
Explanation:
Using Right angle triangle
Let Velocity of ballon&hawk be VHB represent the height of the triangle.
Let Velocity of balloon angle ground be VBG represent adjacent of the triangle.
Let Velocity of hawk and ground BE VHG represent the hypothesis.
Theta = opp/Adj= VHB/VBG
using pythagorean
VHG= SQRT(VHB^2+VBG^2)
VHG= sqrt(2^2+6^2)
VHG= sqrt(4+36)
VHG= 6.325m/s
Tan theta= 2/6
Tan theta =0.3333
Tan^-1 0.3333=18.43°
There is a direct relationship between between an object's weight and its mass.
Weight is actually the force of gravity on the object's mass and it can be calculated by multiplying the object's mass and the acceleration due to gravity. The higher the mass, the higher the weight and the lower the mass, the lower the object's weight. Thus, the relationship is direct.
Responder:
A) ω = 565.56 rad / seg
B) f = 90Hz
C) 0.011111s
Explicación:
Dado que:
Velocidad = 5400 rpm (revolución por minuto)
La velocidad angular (ω) = 2πf
Donde f = frecuencia
ω = 5400 rev / minuto
1 minuto = 60 segundos
2πrad = I revolución
Por lo tanto,
ω = 5400 * (rev / min) * (1 min / 60s) * (2πrad / 1 rev)
ω = (5400 * 2πrad) / 60 s
ω = 10800πrad / 60 s
ω = 180πrad / seg
ω = 565.56 rad / seg
SI)
Dado que :
ω = 2πf
donde f = frecuencia, ω = velocidad angular en rad / s
f = ω / 2π
f = 565.56 / 2π
f = 90.011669
f = 90 Hz
C) Periodo (T)
Recordar T = 1 / f
Por lo tanto,
T = 1/90
T = 0.0111111s
