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3 years ago

4. Water is flowing at 12m/s in a horizontal pipe under a pressure of 600kpa

Physics

1 answer:

shusha [124]3 years ago

4 0

Answer:

a. 192 m/s

b. -17,760 kPa

Explanation:

First let's write the flow rate of the liquid, using the following equation:

Q = A*v

Where Q is the flow rate, A is the cross section area of the pipe (A = pi * radius^2) and v is the speed of the liquid. The flow rate in both parts of the pipe (larger radius and smaller radius) needs to be the same, so we have:

A1*v1 = A2*v2

pi * 0.02^2 * 12 = pi * 0.005^2 * v2

v2 = 0.02^2 * 12 / 0.005^2

v2 = 192 m/s

To find the pressure of the other side, we need to use the Bernoulli equation: (600 kPa = 600000 N/m2)

P1 + d1*v1^2/2 = P2 + d1*v2^2/2

Where d1 is the density of the liquid (for water, we have d1 = 1000 kg/m3)

600000 + 1000*12^2/2 = P2 + 1000*192^2/2

P2 = 600000 + 72000 - 1000*192^2/2

P2 = -17760000 N/m2 = -17,760 kPa

The speed in the smaller part of the pipe is too high, the negative pressure in the second part means that the inicial pressure is not enough to maintain this output speed.

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X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

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3 years ago

A sinusoidal transverse wave of amplitude ym = 8.4 cm and wavelength = 5.3 cm travels on a stretched cord. Find the ratio of the

Scilla [17]

Answer:

The ratio is 9.95

Solution:

As per the question:

Amplitude, $y_{m} = 8.4\ cm$

Wavelength, $\lambda = 5.3\ cm$

Now,

To calculate the ratio of the maximum particle speed to the speed of the wave:

For the maximum speed of the particle:

$v_{m} = y_{m}\times \omega$

where

$\omega = 2\pi f$ = angular speed of the particle

Thus

$v_{m} = 2\pi fy_{m}$

Now,

The wave speed is given by:

$v = f\lambda$

Now,

The ratio is given by:

$\frac{v_{m}}{v} = \frac{2\pi fy_{m}}{f\lambda}$

$\frac{v_{m}}{v} = \frac{2\pi \times 8.4}{5.3} = 9.95$

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3 years ago

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kumpel [21]

Answer:

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Explanation:

CH3COOH is basically vinegar (aka acetic acid)

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2 years ago

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Kipish [7]

Answer:

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Explanation:

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Noble gases are characterized as unreactive atoms, and this is associated with the fact that they have a complete valence shell, it means that they have eight electrons on it (they follow the octet rule).

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