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belka [17]
3 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]3 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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IgorLugansk [536]
<span>From the problem alone we can say that the dart and the block of wood combined into a single object moving together at the end. With that clue we know that the collision is an inelastic collision. The formula of an inelastic collision is:

m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}

First let us sort out our given:Mass should be in kg to get the proper answer. Now let's assign m1 as the mass of the dart and m2 as the mass of the block. 
m1 = 5.5g

5.5g x \frac{1kg}{1000g}= 0.0055kg

m2 = 22.6g

22.6g x \frac{1kg}{1000g}= 0.0226kg

So now we settled that we can set our given as:
M1 = .0055 kg
v1i = ?
M2 = 0.0226 kg
v2i = 0 m/s
dx = 2.5 m
dy = -1.5 m

Now you can see that we have 2 unknowns: v1i and vf. We need the vf to solve for the initial velocity of the dart or object 1. We have other given to consider, so we can make use of that to get our missing vf. 

Now, vf is the horizontal velocity after the collision. We do this by first using the equations for projectiles considering that we have an x and y dimension to consider. We use the y dimension to get the x. 
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dy = -1.5 m 

a = 9.8m/s^2

viy = 0 (take note that the initial vertical velocity is 0)

t = ?

<span>We can use the UAM equations to solve for the time in the y-dimension (vertical) to get the horizontal velocity. 

dy = v_{iy}t +  \frac{1}{2} at^{2}</span>

1.5 = (0)t+\frac{1}{2} (9.8)t^{2}

<span>1.5 = \frac{1}{2} (9.8)t^{2}

\frac{(2)(1.5)}{9.8}=t^{2}

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\sqrt{0.3061} = \sqrt{t^{2}

0.553s = t

Now using this, we can get the horizontal (x-dimension) velocity using the formula:
v_{x} =d_{x}t and our given earlier for the horizontal distance is 2.5m and we solved for time 0.553s. Let's put that into our equation:
v_{x} =d_{x}t
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Now we finally have our vf or velocity after the collision. Now let's get back to the equation.

m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}

From this we can derive the equation for v1i by isolating it. 

v_{1i}= \frac{((m_{1}+m_{2})v_{f})-(m_{2}v_{2i})}{m_{1}}

Now let's put in all our given and what we solved:

v_{1i}= \frac{((0.0055kg+0.0226kg)4.52m/s)-((0.0226kg)0m/s)}{0.0055kg}

v_{1i}= \frac{(0.0281kg)4.52m/s)}{0.0055kg}

v_{1i}= \frac{0.127012kg.m/s}{0.0055kg}

v_{1i}= 23.09m/s

The initial speed of the dart is 23.09 m/s or 23.10 m/s.</span>
7 0
3 years ago
How much energy (in kW-h) does a 900 Watt stove use in a week if it is used for 1.5 hours each day?
ioda

Answer:

9.45 kWh

Explanation:

Energy = Power × time

E = 900 W × (1.5 h/day × 7 day)

E = 9450 Wh

E = 9.45 kWh

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