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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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Find the magnitude of the free-fall acceleration at the orbit of the Moon (a distance of 60RE from the center of the Earth with
Ede4ka [16]

Answer:

The magnitude of the free-fall acceleration at the orbit of the Moon is 2.728\times 10^{-3}\,\frac{m}{s^{2}} (\frac{2.784}{10000}\cdot g, where g = 9.8\,\frac{m}{s^{2}}).

Explanation:

According to the Newton's Law of Gravitation, free fall acceleration (g), in meters per square second, is directly proportional to the mass of the Earth (M), in kilograms, and inversely proportional to the distance from the center of the Earth (r), in meters:

g = \frac{G\cdot M}{r^{2}} (1)

Where:

G - Gravitational constant, in cubic meters per kilogram-square second.

M - Mass of the Earth, in kilograms.

r - Distance from the center of the Earth, in meters.

If we know that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.972\times 10^{24}\,kg and r = 382.26\times 10^{6}\,m, then the free-fall acceleration at the orbit of the Moon is:

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(382.26\times 10^{6}\,m)^{2}}

g = 2.728\times 10^{-3}\,\frac{m}{s^{2}}

6 0
3 years ago
The speed of a tennis ball that is served is 73.14 m/s. During a serve, the ball typically starts from rest and is in contact wi
Serga [27]

Answer: Third option

F = 250w

Explanation:

The impulse can be written as the product of force for the time interval in which it is applied.

I = F (t_2-t_1)

You can also write impulse I as the change of the linear momentum of the ball

I = mv_2 -mv_1

So:

F (t_2-t_1) = mv_2 -mv_1

We want to find the force applied to the ball. We know that

(t_2-t_1) = 30 milliseconds = 0.03 seconds

The initial velocity v_1 is zero.

The final speed v_2 = 73.14\ m / s

So

F * 0.03 = 73.14m

F * 0.03 = 73.14m\\\\F=\frac{73.14m}{0.03}\\\\F=2438m

We must express the result of the force in terms of the weight of the ball.

We divide the expression between the acceleration of gravity

g = 9.8\ m / s ^ 2

F=\frac{2438m*g}{g},\ \ m*g=w\\\\g=9.8\ m/s^2\\\\F=\frac{2438w}{9.8}\\\\F=249w

The answer is the third option

3 0
3 years ago
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