What happens to the rate of deposition as a river gains velocity?

Normally, jet engines push air out the back of the engine, resulting in forward thrust, but commercial aircraft often have thrus

ANEK [815]

Answer:

When the ejected air is moving in the downward direction then the thrust force acts in the upward direction, due to reversal thrust, the jets can take off vertically without needing a runway this way.

Explanation:

Newton’s third law motion states that for every action there will be an equal and opposite reaction.

Thrust reversal is also known as reverse thrust. It acts opposite to the motion of the aircraft by providing the deceleration.

Commercial aircraft moves the ejected air in the forward direction means that the thrust will acts opposite to the motion of the aircraft that is backward direction due to thrust reversal. This thrust force might be used to decelerate the craft.

Uses of thrust reversal in practice:

When the ejected air is moving forward direction then the thrust force moving backward direction due to reversal thrust the speed of the craft slows down.

When the ejected air is moving in the downward direction then the thrust force acts in the upward direction, due to reversal thrust, the jets can take off vertically without needing a runway this way.

6 0

3 years ago

A 2 feet piece of wire is cut into two pieces and once a piece is bent into a square and the other is bent into an equilateral t

SpyIntel [72]

Answer:

Explanation:

Let x ft be used to make square and 2-x ft be used to make equilateral triangle.

each side of square = x/4

area of square = ( x /4 )²

Each side of triangle

= (2-x) /3

Area of triangle = 1/2 (2-x)²/9 sin 60

= √3 / 36 x (2-x)²

Total area

A = ( x /4 )² +√3 / 36 (2-x)²

For maximum area

dA/dx = 0

1/16( 2x ) -√3 / 36 x2(2-x) = 0

x / 8 - √3(2-x)/ 18 = 0

x / 8 - √3/9 + √3/18 x = 0

x ( 1/8 + √3/18 ) = √3/9

x(.125 +.096 ) = .192

x = .868 ft

3 0

3 years ago

water flows through a horizontal pipe with a cross-sectional area of 4m^2 at a speed of 5m/s with a pressure of 300,000pa at poi

LenKa [72]

The velocity at point B is 10 m/s with a pressure of 262500 Pa

<h3>Bernoulli equation</h3>

According to the continuity equation:

A₁V₁ = A₂V₂

Where A is the area and V is velocity, ρ = density of water = 1000 kg/m³

Hence:

4(5) = 2(V₂)

V₂ = 10 m/s

Using Bernoulli equation:

$P_1+\rho gh_1+\frac{1}{2}\rho V_1^2= P_2+\rho gh_2+\frac{1}{2}\rho V_2^2\\\\Hence:\\\\300000+\rho gh+(0.5)*1000*5^2=P_2+\rho gh+(0.5)*1000*10^2\\\\P_2=262500\ Pa\\$