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lora16 [44]
3 years ago
6

View the picture below and then correctly answer the questions using the following words: Temperate Zone, Tropical Zone, Polar Z

one, Polar Winds, Westerly Winds, Trade Winds.
Use the following words to answer the 5 questions below. Temperate Zone, Tropical Zone, Polar Zone, Polar Winds, Westerly Winds, Trade Winds.

1. What winds blow in the zone labeled "B"?

2. What winds blow in the zone labeled "C"?

3. What is the name of the zone labeled "A"?

4. The winds blow in the zone labeled "A"?

5. What is the name of the zone labeled "B"?
Engineering
2 answers:
Nadusha1986 [10]3 years ago
8 0

Can you provide the picture? Thanks !

Otrada [13]3 years ago
7 0
C ITS DEFINITELY C C C C
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Which is one of the aspects in PR game marketing?
mr_godi [17]
C, I took the test already.
7 0
3 years ago
Read 2 more answers
Why the velocity potential Φ(x,y,z,t) exists only for irrotational flow
Black_prince [1.1K]

Answer:

\omega_y,\omega_x,\omega_Z  all are zero.

Explanation:

We know that if flow is possible then it will satisfy the below equation

\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0

Where u is the velocity of flow in the x-direction ,v is the velocity of flow in the y-direction and w is the velocity of flow in z-direction.

And velocity potential function \phi given as follows

 u=-\frac{\partial \phi }{\partial x},v=-\frac{\partial \phi }{\partial y},w=-\frac{\partial \phi }{\partial z}

Rotationality of fluid is given by \omega

\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\omega_Z

\frac{\partial v}{\partial z}-\frac{\partial w}{\partial y}=\omega_x

\frac{\partial w}{\partial x}-\frac{\partial u}{\partial z}=\omega_y

So now putting value in the above equations ,we will find

\omega =\frac{\partial \phi }{\partial x},u=\frac{\partial \phi }{\partial x},

\omega_y=\dfrac{\partial^2 \phi }{\partial z\partial x}-\dfrac{\partial^2 \phi }{\partial z\partial x}

So \omega_y=0

Like this all \omega_y,\omega_x,\omega_Z all are zero.

That is why  velocity potential flow is irroational flow.

5 0
4 years ago
Water from a stationary nozzle impinges on a moving vane with turning angle θ = 120. The vane moves away from the nozzle with co
Tems11 [23]

Answer:

The force that must be applied to maintain the vane speed constant is 2771.26 N

Explanation:

Given;

turning angle of the vane = 120°

control volume velocity, u = 10 m/s

absolute velocity, v = 30 m/s

nozzle area = 0.004 m²

The force acting on the vane has horizontal and vertical components:

Based on Reynolds general control volume system;

The horizontal force component of the system, ∑Fₓ = ρW²A(1-cosθ)

where;

ρ is the density of water = 1000 kg/m³

W is the relative velocity = Absolute velocity - control volume velocity

W = v - u

    = 30 - 10 = 20m/s

∑Fₓ = ρW²A(1-cosθ) = 1000 x 20² x 0.004 (1 - cos 120) = 2400 N

The vertical force component of the system, ∑Fy = ρW²A(sinθ)

∑Fy = ρW²A(sinθ) = 1000 x 20² x 0.004 x sin(120) = 1385.6 N

The magnitude of the force applied = \sqrt{F_x^2 + F_y^2}

F = \sqrt{2400^2 + 1385.6^2} = 2771.26 \ N

The force that must be applied to maintain the vane speed constant is 2771.26 N

6 0
3 years ago
Read 2 more answers
3. Technician A says passive permanent
Angelina_Jolie [31]

Answer:

Both technician A and B

Explanation:

Passive permanent  magnet ABS wheel speed sensors produce an  A/C voltage signal. Wheel speed sensors are a necessary ABS component and sensor input. It is used to inform the ABS control module of rotational wheel speed. A passive sensor creates an AC signal that changes frequency as the wheel changes speed. Moreover, input  from wheel speed sensors are used for anti- lock brake, electronic traction control, and  electronic stability control systems. Therefore, both technicians are correct.

6 0
4 years ago
A car starts out from rest (zero velocity) at an elevation of 500 m and drives up a hill to reach a final elevation of 2000m and
natta225 [31]

Answer:29,930 kJ

Explanation:

Given

Car starts with an initial elevation of 500 m and drives up a hill to reach a final elevation of 2000 m

Final velocity (V)=20 m/s

Energy of car increases by 100 kJ

mass of car(m)=2000 kg

Total Energy =\Delta PE+\Delta KE+\Delta U

\Delta PE=mg(\Delta h)=2000\times 9.81\times (2000-500)

\Delta PE=29,430 kJ

\Delta KE=m\frac{v_2^2-v_1^2}{2}

\Delta KE=2000\times \frac{20^2-0^2}{2}

\Delta KE=400 kJ

\Delta U=100 kJ

Total Energy=29,430+400+100=29,930 kJ

4 0
4 years ago
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