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Leno4ka [110]
3 years ago
13

A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat

es horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 27.9 m/s2 with a beam of length 5.21 m , what rotation frequency is required
Physics
1 answer:
slava [35]3 years ago
8 0

Answer: the angular frequency is 2.31 rad/s

Explanation:

The data we have is:

Radial acceleration A = 27.9 m/s^2

Beam length r = 5.21m

The radial acceleration is equal to the velocity square divided the radius of the circle (the lenght of the beam in this case)

And we can write the velocity as:

v = w*r where r is the radius of the circle, and w is the angular frequency.

w = 2pi*f

where f is the "normal" frequency.

So we have:

A = (v^2)/r = (r*w)^2/r = r*w^2

We can replace the values and find w.

27.9m/s^2 = 5.21m*w^2

√(27.9/5.21) = w = 2.31 rad/s

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Acceleration in a velocity vs time graph is just the slope at that point. The reason for that is because the definition of acceleration is the change in velocity per unit of time. In this case we want instantaneous, which is the derivative or tangent line at that point.

At 3s we can see the slope is 0, so that means his acceleration is zero. That means he was moving at a constant velocity

At 5s we can see that the slope is negative. And from 5s to 6s the change in velocity is -5m/s^2

At 7s we can see the slope is very positive. And from 7s to 8s the change in velocity is +15m/s^2

And again, at 9s the slope is 0 so his acceleration is also zero. He’s moving at a constant velocity

If you take the integral of a velocity vs time graph, you get position. So the area underneath a velocity vs time graph is the distance traveled. Anything below the x axis is considered negative distance. We need to take the area of a triangle and the area of two rectangles to find the distance.

So, let’s do the two rectangles first. From 8s to 9s it is a width of 1 and a length of 40. So the area would be 40 meters. Let’s do the second rectangle. From 7s to 8s it is a width of 1. Then the length goes up to 25. So the area is 25 meters.

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6 0
3 years ago
A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.4 A runs through the wire
Deffense [45]

Answer:

\vec{F}=0.40176 N \hat{k}

Explanation:

To calculate the force we need to use this equation

\vec{F}=\int\limits^L_0 {i(\vec{dl}\times\vec{B})}

where L is the total length of the wire

So in this case the small element of current is

\vec{dl} = dx \hat{i}

Because x is the direction of the current flow.

As is said in the problem B is such that

\vec{B} = B \hat{j} = 0.62\hat{j} [ T]

so to use the equation above we first calculate the following cross product:

\vec{dl}\times\vec{B}=dx \hat{i}\times B \hat{j} = Bdx\hat{k}

so the force:

F = \int\limits^L_0 {i(\vec{dl}\times\vec{B})}=\int\limits^L_0{iBdx\hat{k}}

So here we use the fact that B=0 in any point of the x axis that is not x^{'}=0.27 [m], that means that we only need to do the integration between a very short distant behind the point x^{'}=0.27 [m] and a very short distant after that point, meaning:

\vec{F}= \lim_{h \to 0}{\int\limits^{x^{'}+h}_{x^{'}-h}{iBdx\hat{k}} }

so is the same as evaluating iBx at x=x^{'}

that is:

2,4 A * 0,62 T * 0,27 m \hat{k}

2,4 A * 0,62 (\frac{Kg}{A s^{2}}) * 0,27 m \hat{k}

2,4*0,62*2,7 ( \frac{ kgm }{ s^{2} } ) \hat{k}

\vec{F}=0.40176 N \hat{k}

5 0
3 years ago
Electricity is the _____________ of electrons, because electrons _____________ from atom to atom.
Oksana_A [137]
Electricity is the FLOW of electrons, because electrons PASS from atom to atom
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Amiraneli [1.4K]

Answer:

B

Explanation:

transferred

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4 years ago
Determine the speed
Paraphin [41]
Calculating the average speed is simple using the formula <span>speed = distance/time</span>
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