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RoseWind [281]
3 years ago
6

A jetliner, traveling northward, is landing with a speed of 71.9 m/s. Once the jet touches down, it has 675 m of runway in which

to reduce its speed to 11.3 m/s. Compute the average acceleration (magnitude and direction) of the plane during landing (take the direction of the plane's motion as positive).
Physics
1 answer:
Leviafan [203]3 years ago
5 0

Answer:

The value is  a =  -3.7 \  m/s^2

Explanation:

From the question we are told that

   The  landing speed is  u =  71.9 \  m/s

   The  distance traveled is  d =  675 \  m

    The velocity it is reduced to is  v  =  11.3 \  m/s

   

Generally the average acceleration is mathematically represented as

      a =  \frac{ v^2  -  u^2 }{ 2 * d }

=>  a =  \frac{ 11.3^2 - 71.9^2 }{ 2 * 675 }

=>   a =  -3.7 \  m/s^2

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A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a di
Vikentia [17]

Answer:

option B

Explanation:

given,

Force exerted by the hydraulic jack piston = F₁ = 250 N

diameter of piston, d₁ = 0.02 m

                                r₁ = 0.01 m

diameter of second piston,  d₂ = 0.15 m

                                r₂ = 0.075 m

mass of the jack to lift = ?

now,

    \dfrac{F_1}{A_1} =\dfrac{F_2}{A_2}

    \dfrac{250}{\pi r_1^2} =\dfrac{F_2}{\pi r_2^2}

    \dfrac{250}{0.01^2} =\dfrac{F_2}{0.075^2}

    F_2= \dfrac{250}{0.01^2}\times {0.075^2}

               F₂ = 14062.5 N

F = m g

m = \dfrac{F}{g}

m = \dfrac{14062.5}{9.8}

m = 1435 Kg

hence, the correct answer is option B

5 0
3 years ago
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