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2 years ago

If the Test Section area has an air speed of 10 m/s, what would be the air speed at the Air Intake area

Physics

1 answer:

FrozenT [24]2 years ago

4 0

The speed of air at the intake area is $\frac{10A_2}{A_1}$ (m/s).

<h3 /><h3>Continuity equation</h3>

The continuity equation is used to determine the flow rate at different sections of a pipe or fluid conduit.

The continuity equation is given as;

$A_1v_1 = A_2 v_2$

Let the intake area = A₁
Let the velocity of air in intake area = v₁
Let the area of the test section = A₂
Velocity of air in test section, v₂ = 10 m/s

The speed of air at the intake area is calculated as follows;

$A_1 v_1 = A_2 v_2\\\\
v_1 = \frac{A_2 v_2}{A_1} \\\\
v_1 = \frac{10A_2}{A_1}$

Learn more about continuity equation here: brainly.com/question/14619396

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A stereo speaker produces a pure \"c\" tone, with a frequency of 262 hz. what is the period of the sound wave produced by the sp

jasenka [17]

Answer:

3.82 ms

Explanation:

The period of a wave is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f is the frequency.

In this problem, f = 262 Hz, so the period if this sound wave is

$T=\frac{1}{262 Hz}=3.82\cdot 10^{-3} s=3.82ms$

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3 years ago

The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental

pav-90 [236]

A) $5.0\cdot 10^{-11} m$

The energy of an x-ray photon used for single dental x-rays is

$E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J$

The energy of a photon is related to its wavelength by the equation

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

Re-arranging the equation for the wavelength, we find

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m$

B) $2.0\cdot 10^{-11} m$

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

$E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J$

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m$

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2 years ago

Calculate the mass of an ice block of volume 12m³. the density of the ice is 920kg/m³.

beks73 [17]

Mass = density • volume

so (12)(920)(kg/m^3•m^3)

11040 kg

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2 years ago

A ball of mass m falls vertically, hits the floor with a speed ui , and rebounds with a speed u f . what is the magnitude of the

Sholpan [36]

Answer:

Change of momentum = M (Vf - (-Vi)) where V represents the scalar speeds of the ball or

I = M (ui + uf) and I is the impulse ΔM V = I Force = Change in Momentum

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1 year ago

A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity incr

____ [38]

Answer:

The number of revolutions is 44.6.

Explanation:

We can find the revolutions of the wheel with the following equation:

$\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}$

Where:

$\omega_{0}$ : is the initial angular velocity = 13 rad/s

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

$\omega_{f} = \omega_{0} + \alpha t$

Where:

$\omega_{f}$ : is the final angular velocity = 57 rad/s

$\alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2}$

Hence, the number of revolutions is:

$\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev$

Therefore, the number of revolutions is 44.6.

I hope it helps you!

4 0

2 years ago