Explanation:
The critical velocity is that velocity of liquid flow, up to which its flow is streamlined (laminar)& above which its flow becomes turbulent. It's denoted by Vc & it depends upon: Coefficient of viscosity of liquid (η) Density of liquid. Radius of the tube.
Answer:
0.5 m/s²
Explanation:
according to Newton's second law, we are goven a relationship between force, mass and acceleration, with the formula:
F = m×a
F for force
m for mass
a for acceleration
we use the given data and get:
20 = 40×a
we find a=20/40=0.5m/s²
1.) because then people can evacuate the area in the path<span> of the hurricane.
2.) </span><span>At higher altitudes, water vapor starts to condense into clouds and rain, releasing heat that warms the surrounding air, Which makes it rise as well. Warmer waters feed more energetic storms.
3.) </span> <span>A hurricane starts off as a series of thunderstorms which intensify as it moves over the warm and humid sea. The humidity is at a constant level and so it continues to grow over the sea. Any kind of decrease or increase in humidity can change the strength of a hurricane.
4.) </span><span>Actually, tropical cyclones need weak winds. If the atmospheric winds are even remotely strong, they will act to cut back the system and prevent the convection from wrapping around the center.
</span><span>Annndd...
5.) That hard to tell, it could be too much. Though I am going to go with yes. Cyclones need weak winds and good amount humidity.</span><span>
</span>
Answer:
T₂ = 123.9 N, θ = 66.2º
Explanation:
To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.
The tension T1 = 100 N, we create a reference frame centered on the pole
X axis
T₁ₓ -
= 0
T_{2x}= T₁ₓ
Y axis y
T_{1y} + T_{2y} - 200N = 0
T_{2y} = 200 -T_{1y}
let's use trigonometry to find the component of the stresses
sin 60 = T_{1y} / T₁
cos 60 = t₁ₓ / T₁
T_{1y} = T₁ sin 60
T1x = T₁ cos 60
T_{1y}y = 100 sin 60 = 86.6 N
T₁ₓ = 100 cos 60 = 50 N
for voltage 2 it is done in the same way
T_{2y} = T₂ sin θ
T₂ₓ = T₂ cos θ
we substitute
T₂ sin θ= 200 - 86.6 = 113.4
T₂ cos θ = 50 (1)
to solve the system we divide the two equations
tan θ = 113.4 / 50
θ = tan⁻¹ 2,268
θ = 66.2º
we caption in equation 1
T₂ cos 66.2 = 50
T₂ = 50 / cos 66.2
T₂ = 123.9 N