An airplane flies from San Francisco to Washington DC at an air speed of 800 km/hr. Assume Washington is due east of San Francisco at a distance of 6000 km. Use a Cartesian system of coordinates centered at San Francisco with Washington in the positive x-direction. At cruising altitude, there is a cross wind blowing from north to south of 100 km/hr.

Required:
a. What must be the direction of flight for the plane to actually arrive in Washington?
b. What is the speed in the San Francisco to Washington direction?
c. How long does it take to cover this distance?
d. What is the time difference compared to no crosswind?

Answer:

Explanation:

The value of a will be zero as it is provided that the particle is on the x-axis.

Calculate the velocity of particles along x-axis.

[tex]{\bf{V}} = 2{x^2}t{\bf{\hat i}} + [4y(t - 1) + 2{x^2}t]{\bf{\hat j}}{\rm{ m/s}}[/tex]

Substitute 0 for y.

[tex]\begin{array}{c}\\{\bf{V}} = 2{x^2}t{\bf{\hat i}} + \left( {4\left( 0 \right)\left( {t - 1} \right) + 2{x^2}t} \right){\bf{\hat j}}{\rm{ m/s}}\\\\ = 2{x^2}t{\bf{\hat i}} + 2{x^2}t{\bf{\hat j}}{\rm{ m/s}}\\\end{array}[/tex]

Here,

[tex]A = 2{x^2}t \ \ and\ \ B = 2{x^2}t[/tex]

Calculate the magnitude of vector V .

[tex].\left| {\bf{V}} \right| = \sqrt {{A^2} + {B^2}}[/tex]

Substitute

[tex]2{x^2}t \ \ for\ A\ and\ 2{x^2}t \ \ for \ B.[/tex]

[tex]\begin{array}{c}\\\left| {\bf{V}} \right| = \sqrt {{{\left( {2{x^2}t} \right)}^2} + {{\left( {2{x^2}t} \right)}^2}} \\\\ = \left( {2\sqrt 2 } \right){x^2}t\\\end{array}[/tex]

The velocity of the fluid particles on the x-axis is [tex]\left( {2\sqrt 2 } \right){x^2}t{\rm{ m/s}}[/tex]

Calculate the direction of flow.

[tex]\theta = {\tan ^{ - 1}}\left( {\frac{B}{A}} )[/tex]

Here, θ is the flow from positive x-axis in a counterclockwise direction.

Substitute [tex]2{x^2}t[/tex] as A and [tex]2{x^2}t[/tex] as B.

[tex]\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{2{x^2}t}}{{2{x^2}t}}} \right)\\\\ = {\tan ^{ - 1}}\left( 1 \right)\\\\ = 45^\circ \\\end{array}[/tex]

The direction of flow is [tex]45^\circ[/tex] from the positive x-axis.

Answer:

The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s

Explanation:

Assuming the two oils are Newtonian fluids.

From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.

τ = μ (∂v/∂y)

There are oils above and below the plate, so we can write this expression for the both cases.

τ₁ = μ₁ (∂v/∂y)

τ₂ = μ₂ (∂v/∂y)

dv = 0.3 m/s

dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)

τ₁ = μ₁ (0.3/0.03) = 10μ₁

τ₂ = μ₂ (0.3/0.03) = 10μ₂

But the shear stress on the plate is given as 29 N per square meter.

τ = 29 N/m²

But this stress is a sum of stress due to both shear stress above and below the plate

τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29

But it is also given that one viscosity is twice the other

μ₁ = 2μ₂

10μ₁ + 10μ₂ = 29

10(2μ₂) + 10μ₂ = 29

30μ₂ = 29

μ₂ = (29/30) = 0.967 Pa.s

μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s

Hope this Helps!!!

Question:Technician A say's that The most two-stroke engines have a pressure type lubrication system. Technician be says that four stroke engines do not require the mixing of oil with gasoline . Which of them is correct ?

Answer: Technician  B is   correct

Explanation: Two types of engines exist , the two stroke (example, used in chainsaws)  is a type of engine that uses two strokes--a compression stroke and a return stroke to produce power in a crankshaft combustion cycle and the  four stroke engines(eg lawnmowers) which  uses four strokes,  2-strokes during  compression and exhaustion accompanied by 2 return strokes for each of the initial process to produce power in a combustion cycle.

While a 2 stroke system engine, requires mixing of oil and fuel to the crankshaft before  forcing  the mixture into the cylinder and do not require a pressurized system.  The 4 stroke system uses a splash and pressurized system where oil is not mixed with gasoline but drawn from the sump and  directed to  the main moving  parts of  crankshaft through its channels.

We can therefore say that Technician A is wrong while Technician B is  correct

Answer:

height ≈ 60.60 m

Explanation:

The surveyor is trying to find the height of the hill . He takes a sight on the top of the hill and finds the angle of elevation is 40°. The distance from the hill where he measured the angle of elevation of 40° is not known.

Now he moves 150 m on level ground directly away from the hill and take a second sight from this point and measures the angle of elevation as 22°. This illustration forms a right angle triangle. The opposite side of the triangle is the height of the hill. The adjacent side of the triangle which is 150 m is the distance on level ground directly away from the hill.

Using tangential ratio,

tan 22° = opposite/adjacent

tan 22° = h/150

h = 150 × tan 22°

h = 150 × 0.40402622583

h = 60.6039338753

height ≈ 60.60 m

Answer:

To answer this question we assumed that the area units and the thickness units are given in inches.

The number of atoms of lead required is 1.73x10²³.    

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

[tex] V = A*t [/tex]

Where:

A: is the surface area = 160

t: is the thickness = 0.002

Assuming that the units given above are in inches we proceed to calculate the volume:

[tex]V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}[/tex]    

Now, using the density we can find the mass:

[tex] m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g [/tex]

Finally, with the Avogadros number ([tex]N_{A}[/tex]) and with the atomic mass (A) we can find the number of atoms (N):

[tex] N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms [/tex]    

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

Answer:

Exit temperature = 32°C

Explanation:

We are given;

Initial Pressure;P1 = 100 KPa

Cp =1000 J/kg.K = 1 KJ/kg.k

R = 500 J/kg.K = 0.5 Kj/Kg.k

Initial temperature;T1 = 27°C = 273 + 27K = 300 K

volume flow rate;V' = 15 m³/s

W = 130 Kw

Q = 80 Kw

Using ideal gas equation,

PV' = m'RT

Where m' is mass flow rate.

Thus;making m' the subject, we have;

m' = PV'/RT

So at inlet,

m' = P1•V1'/(R•T1)

m' = (100 × 15)/(0.5 × 300)

m' = 10 kg/s

From steady flow energy equation, we know that;

m'•h1 + Q = m'h2 + W

Dividing through by m', we have;

h1 + Q/m' = h2 + W/m'

h = Cp•T

Thus,

Cp•T1 + Q/m' = Cp•T2 + W/m'

Plugging in the relevant values, we have;

(1*300) - (80/10) = (1*T2) - (130/10)

Q and M negative because heat is being lost.

300 - 8 + 13 = T2

T2 = 305 K = 305 - 273 °C = 32 °C

Mathematical modeling aids in technological design by simulating how proposed system might behave. The correct option is 2.

Mathematical modelling describes a real world problem in mathematical terms or in the form of equations. This makes an engineer to discover new features about the problem and designer to alter his design for better function and output.

Mathematical models allow engineers and designers to understand how the proposed model and actual prototype will be produced.

Thus, the correct option is 2.

Learn more about mathematical modelling

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Answer:

In consolidated AB Co 300 shares.

Explanation:

Consolidation is a process in which two different organizations are united. In this question A Co and B Co are consolidated and a new Co names AB Co is formed. The shares of both the companies will be combined and their total share capital will be increased.