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.
a surveyor is trying to find the height of a hill . he/she takes a sight on the top of the hill and find that the angle of elevation is 40°. he/she move a distance of 150 metres on level ground directly away from the hill and take a second sight. from this point the angl.e of elevation is 22°. find the height of the
hill
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
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
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