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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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Membrane walls of living cells have surprisingly large electric fields across them due to separation of ions. (Membranes are dis
Arturiano [62]

Answer:

Voltage, V = 0.0524 volts

Explanation:

Thickness of the membrane, d=7.95\ nm=7.95\times 10^{-9}\ m

Electric field strength, E=6.6\ MV/m=6.6\times 10^6\ V/m

We need to find the voltage across it. The relationship between the voltage, electric field and the distance is given by :

V=E\times d

V=6.6\times 10^6\ V/m\times 7.95\times 10^{-9}

V = 0.0524 volts

So, the voltage across the thick membrane is 0.0524 volts. Hence, this is the required solution.

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3 years ago
One similarity between work and power is that in order to calculate both you must know
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D.) In order to calculate both of them, we must know the "FORCE" on the system.
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3 years ago
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The wavelength of a sound wave in this room is 1.13 m and the frequency is 301 Hz.
Yakvenalex [24]

Answer:

a. 340.13 m/s b. 680.26 m/s c. our wavelength doubles

Explanation:

a. speed of wave, v = fλ were f = frequency = 301 Hz and λ = wavelength = 1.13 m.

v = fλ = 301 Hz × 1.13 m = 340.13 m/s

b. If we double the frequency then f = 2 × 301 Hz = 602 Hz

v = fλ = 602 Hz × 1.13 m = 680.26 m/s

c. If the speed of the wave is still 340.13 m/s, if we cut the frequency in half, then frequency now equals f = 301 Hz/2 = 150.5 Hz.

Since v = fλ,

λ = v/f = 340.13 m/s ÷ 150.5 Hz = 2.26 m.

Since our initial wavelength λ₀ = 1.13 m,

λ/λ₀ = 2.26 m/1.13 m = 2.

So, λ = 2λ₀ our wavelength doubles

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3 years ago
the length of the glass at 25°C is 500cm after gaining some heat the final length of the glass becomes 500.9cm determine the tem
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3 years ago
Given a double slit apparatus with slit distance 2 mm, what is the theoretical maximum number of bright spots that I would see w
katen-ka-za [31]

Answer:

The values is  m_{max} = 8001 \  bright \ spots

Explanation:

From the question we are told that

    The slit distance is  d  =  2 \ mm  =  2*10^{-3} \ m

    The  wavelength is  \lambda =  500 \ nm =  500 *10^{-9} \ m

At the first half of the screen from the central maxima

   The number of bright spot according to the condition for constructive interference is  

          n  =  \frac{d *  sin (\theta )}{\lambda}

For maximum number of spot \theta =  90^o

So  

       n  =  \frac{2*10^{-3} *  sin (90 )}{500 *10^{-9}}

        n  =4000

Now for the both sides plus the central maxima  we have

      m_{max} = 2 * n  + 1

substituting values

       m_{max} = 2 *  4000 + 1

       m_{max} = 8001 \  bright \ spots

   

6 0
4 years ago
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