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GaryK [48]
3 years ago
13

2. An airplane traveling north at 220. meters per second encounters a 50.0-meters-per-second crosswind

Physics
1 answer:
Alex777 [14]3 years ago
5 0

The resultant speed of the plane  is (3) 226 m/s

Why?

We can calculate the resultant speed of the plane by using the Pythagorean Theorem since both speeds are perpendicular (forming a right triangle).

So, calculating we have:

ResultantSpeed=\sqrt{VerticalSpeed^{2}+HorizontalSpeed^{2}}\\\\ResultantSpeed=\sqrt{(220\frac{m}{s})^{2}+50\frac{m}{s})^{2}

ResulntantSpeed=\sqrt{48400\frac{m^{2} }{s^{2} }+2500\frac{m^{2} }{s^{2} } } \\\\ResultantSpeed=\sqrt{50900\frac{m^{2} }{s^{2} }}=226\frac{m}{s}

Hence, we have that the resultant speed of the plane  is (3) 226 m/s

Have a nice day!

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Answer:

 ∑ τ =0,  L₀ = L_{f}

Explanation:

In a circular turning movement, when the arms are extended and then contracted in two possibilities:

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    In this movement the force is vertical(gravity) and the movement of the center of mass of each arm is vertical, so that the work is the weight value of the arm by the distance traveled by the center of mass.

- Another possibility is that the arms have stuck to the body, in this case the person's muscles perform the force, this force is horizontal and the displacement is the horizontal of the center of mass of the arms from the extended position to the contracted

 

In these movements the torque of the external force is equal for each arm, but in the opposite direction, so they are canceled where a net torque of zero, this causes the angular momentum to be preserved, which changes is the moment of inertia of the system and therefore you must also change the angular velocity to keep your product constant

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                L₀ = L_{f}

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A plane comes in for a landing at a velocity of 80 meters per second west. As it touches down, it decelerates at a constant rate
azamat

Answer:

Answer D : about 1067 meters

Explanation:

There are two steps to this problem:

1) First find the time it takes the plane to stop using the equation for the acceleration:

a=\frac{Vf-Vi}{t}

Where Vf is the final velocity of the plane (in our case: zero )

Vi is the initial velocity of the plane (in our case: 80 m/s)

a is the acceleration (in our case -3 m/s^2 - notice negative value because the velocity is decreasing)

a=\frac{Vf-Vi}{t}\\-3=\frac{0-80}{t}\\t=\frac{-80}{-3} = \frac{80}{3}

with units corresponding to seconds given the quantities involved in the calculation.

2) Second knowing the time it took the plane to stop, now use that time in the equation for the distance traveled under accelerated motion:

Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi= 80 (\frac{80}{3}) +\frac{1}{2} (-3) (\frac{80}{3}) ^{2}=1066.666666...

Where the answer results in units of meters given the quantities used in the calculation.

We round this to 1067 meters

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