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dexar [7]
3 years ago
6

A small, single engine airplane is about to take off. The airplane becomes airborne, when its speed reaches 193.0 km/h. The cond

itions at the airport are ideal, there is no wind. When the engine is running at its full power, the acceleration of the airplane is 2.80 m/s2. What is the minimum required length of the runway
Physics
1 answer:
siniylev [52]3 years ago
4 0

Answer:

513 m

Explanation:

We have;

final speed of the airplane = 193.0 km/h * 1000/3600 = 53.6 m/s

acceleration of the air plane = 2.80 m/s2

initial velocity of the airplane = 0 m/s

length of the runway = distance covered

v^2 = u^2 + 2as

v^2 - u^2 = 2as

s = v^2 - u^2/2a

s = (53.6)^2 - 0^2/ 2 *  2.80

s = 2872.96/ 5.6

s = 513 m

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A 2 kg ball is moving 3 m/s when it starts rolling up a hill.
AURORKA [14]

Answer:

the height reached is = 0.458 [m]

Explanation:

We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:

Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]

Replacing the values on the equation we have:

Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\

This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.

Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\

Now we have:

h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]

In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]

6 0
3 years ago
The famous scientist Galileo Galilei did several experiments with sloping planes, which he rolled metal balls down so that he co
KatRina [158]
Hi there!

The answer would be B. the slope of the plane.

Changing the slope of the plane would show how fast the ball went when Galileo changed the steepness of the slope. If he didn’t change the slopes steepness he would have the same results each time.

Hope this helps !
5 0
3 years ago
There is a reflection at the boundary between two materials if there is a change of __________ at the boundary.
melisa1 [442]
<span>The speed of the light in the materials. Hope this helps!!</span>
8 0
3 years ago
A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t
svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

\omega=-\alpha t

\alpha=-\dfrac{\omega}{t}

\alpha=-\dfrac{50.0}{20.0}

\alpha=-2.5\ rad/s^2

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

\theta=\omega_{0}t+\dfrac{1}{2}\alpha t

Put the value into the formula

\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2

\theta=500\ rad

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

\vec{\tau}=\vec{r}\times\vec{f}

\tau=r\times f\sin\theta

Put the value into the formula

\tau=2.5\times4\times 9.8\sin60

\tau=84.87\ N-m

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

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3 years ago
A continental polar air mass forms in ____
Sladkaya [172]
I believe it is A bit not completely sure
7 0
3 years ago
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