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belka [17]
4 years ago
10

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

t reversers that can change the direction of the ejected air, sending it forward. How does this affect the direction of thrust? When might these thrust reversers be useful in practice?
Physics
1 answer:
ANEK [815]4 years ago
6 0

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.

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A level curve on a country road has a radius of 150 m. What is the maximum speed at which this curve can be safely negotiated on
SSSSS [86.1K]

The maximum speed of the car is 24.3 m/s

Explanation:

For a car moving along an unbanked turn, the frictional force provides the centripetal force required to keep the car in circular motion. Therefore, we can write:

\mu mg = m\frac{v^2}{r}

where the term on the left is the frictional force while the term on the right is the centripetal force, and where

\mu=0.40 is the coefficient of friction between the tires and the road

m is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

v is the speed of the car

r = 150 m is the radius of the curve

Solving for v, we find the (maximum) speed at which the car can move along the turn:

v=\sqrt{\mu gr}=\sqrt{(0.40)(9.8)(150)}=24.3 m/s

For speed larger than this value, the frictional force is no longer enough to keep the car along the turn.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0
3 years ago
WILL MARK BRAINLIST!!! only for correct answer
adelina 88 [10]

Answer:

I think the answer is C. for 4. and then B. for 5.

4 0
3 years ago
Calculate the period (T) of uniform circular motion if the velocity is 40.0 m/s and centripetal acceleration is 20.0 m/s2.
kirill [66]
T is the time for a whole round.

centripetal acceleration = V^2/R,

20 = 40^2 / R, find R = 40^2/20 = 40*40/20 = 80 m, right?

Now, one round is L = 2*pi*R = 2*pi*80 = 160*pi

And T = L/v (distance/speed) = 160*pi/40 = 4*pi seconds, or ~ 12.57 s
3 0
4 years ago
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aleksandr82 [10.1K]

Answer:

I have examined the different parts of blub and I assume the answer is (D)T and U

Hope this helps you

4 0
3 years ago
The ball is launched with total initial energy Einitial = mgho. When it reaches the bottom of the hoops, it now has a potential
NeX [460]

Answer:

  K = m g (h₀ - h_plat)

Explanation:

Let's use energy conservation to solve this problem, write the energy at two points of interest

Initial. Higher

          E_initial = m g h₀

Final. Lower

          E_end = K + Ep

          E_end = ½ m v² + m g h_plat

Energy is conserved

          E_initial = E_end

          m g h₀ = K + m g h_plat

          K = m g (h₀ - h_plat)

4 0
3 years ago
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