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sattari [20]
3 years ago
7

Rainwater draining from a neighborhood street initially travels at 4 ft/s through a pipe with a cross-sectional area of 15.7 ft2

. This pipe connects down the street to another pipe with a cross-sectional area of 65.4 ft2. What would be the speed of the water as it moves through this larger pipe?
Physics
1 answer:
Fudgin [204]3 years ago
6 0

Answer:

The  velocity  is v_2  =  0.96 \ ft/s

Explanation:

From the question we are told that

   The initial speed is  v_1  =  4 \ ft/s

   The  cross -sectional area of the first pipe is  A_1  =  15.7 \ ft

   The  cross -sectional area of the second pipe is A_2 =  65.4 \  ft

Generally from continuity equation we have that

     A_1 * v_1 =  A_2  * v_2

So  

     v_2  =  \frac{A_1 *  v_1  }{A_2 }

=>   v_2  =  \frac{15.7  *  4  }{65.4 }

=>   v_2  =  0.96 \ ft/s

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How does the rotation of a galaxy result in spectral line broadening?
IRISSAK [1]

Answer:

Explanation:

Normal galaxies are made up of stars and (in the case of spiral and irregular galaxies) gas and dust. Their spectra consist of the sum of the spectra of these components.

The optical spectra of normal stars are continuous spectra overlaid by absorption lines (Figure 1). There are two factors to consider when adding up the spectra of a number of stars to produce the spectrum of a galaxy:

Different types of star have different absorption lines in their spectra. When the spectra are added together, the absorption lines are 'diluted' because a line in the spectrum of one type of star may not appear in the spectra of other types.

Doppler shifts can affect all spectral lines. All lines from a galaxy share the red-shift of the galaxy, but Doppler shifts can also arise from motions of objects within the galaxy. As a result, the absorption lines become broader and shallower. We explain below how this Doppler broadening comes about.

HII regions in spiral and irregular galaxies (though not, of course, ellipticals) shine brightly and contribute significantly to the spectrum of the galaxy. The optical spectrum of an HII region consists mainly of emission lines, as in Figure 2. When the spectra of the HII regions and the stars of a galaxy are added together, the emission lines from the HII regions tend to remain as prominent features in the spectrum unless a line coincides with a stellar absorption line. There are Doppler shift effects, however, as described for stellar absorption lines, and hence emission lines too are broadened because of the motion of HII regions within a galaxy.

Box 1: Doppler Broadening

The Doppler effect causes wavelengths to be lengthened when the source is moving away from the observer (red-shifted) and shortened when the source is moving towards the observer (blue-shifted).

Light from an astrophysical source is the sum of many photons emitted by individual atoms. Each of these atoms is in motion and so their photons will be seen as blue- or red-shifted according to the relative speeds of the atom and the observer. For example, even though all hydrogen atoms emit H photons of precisely the same wavelength, an observer will see the photons arrive with a spread of wavelengths: the effect is to broaden the H spectral line - called Doppler broadening.

In general, if the emitting atoms are in motion with a range of speeds Δν along the line of sight to the observer (the velocity dispersion) then the Doppler broadening is given by

where c is the speed of light, and λ is the central wavelength of the spectral line.

Why would the atoms be in motion? An obvious reason is that they are 'hot'. Atoms in a hot gas, for example, will be moving randomly with a range of speeds related to the temperature of the gas. For a gas of atoms of mass m at a temperature T, the velocity dispersion is given by

where k is the Boltzmann constant (1.38 × 10−23 J K−1).

4 0
3 years ago
A car mass 600kg starts from rest moving uniform acceleration 0.2 m/s^2 after 60 seconds collides with stationary pick up van of
nlexa [21]

Answer:

The given phenomenon  supports the Principle of Conservation of momentum.

Explanation:

law of conservation of momentum  

Initial Momentum = Final Momentum

So, first we calculate initial momentum of the system:

Initial Momentum  = m1*u1 + m2*u2

and we are given

m1 = mass of car = 600 kg

m2 = mass of van = 400 kg

u1 = Initial Speed of Car  

to find the initial speed we use equation of motion

Vf = Vi + at

Vf = 0 m/s + (0.2 m/s²)(60 s)

Vf = u₁ = 12 m/s  

u₂ = Initial Speed of Van = 0 m/s

Therefore,

Initial Momentum  = (600 kg)(12 m/s) + (400 kg)(0 m/s)

Initial Momentum  = 7200 Ns   .........(1)

final momentum = m₁v₁ + m₂v₂

where,

v₁ = v₂ = final speed of car+van (both locked ) = 7.2 m/s

Therefore,

Final Momentum = (600 kg)(7.2 m/s) + (400 kg)(7.2 m/s)

Final Momentum = 7200 Ns   -------- (2)

on comparing (1) and (2)

Initial momentum = Final Momentum

Hence, the phenomenon of the system supports the principle of conservation of momentum.

6 0
3 years ago
A man pushes his lawnmower with a velocity of +0.75 m/s relative to the ground. A girl rides by on her bike with a velocity of +
SashulF [63]

Answer:

B. +5.75 m/s

Explanation:

When there are two bodies, a and b, whose velocities measured by a third observer (in this case, the ground) are V_a and V_b respectively, the relative velocity of B with respect to A is given by:

V_{ba}=V_b-V_a

Thus, the velocity of the girl relative to the lawnmower is:

V_{ba}=6.5\frac{m}{s}-0.75\frac{m}{s}\\V_{ba}=5.75\frac{m}{s}

8 0
3 years ago
Read 2 more answers
A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.
andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}

Energy

\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})

m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}

Where:

m_{1}, m_{2} - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

v_{1,o}, v_{2,o} - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

v_{1}, v_{2} - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If m_{1} = 0.400\,kg, m_{2} = 0.900\,kg, v_{1,o} = +5.86\,\frac{m}{s}, v_{2,o} = 0\,\frac{m}{s}, the system of equation is simplified as follows:

2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}

13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}

Let is clear v_{1,f} in first equation:

0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}

v_{1,f} = 5.86-2.25\cdot v_{2,f}

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}

13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}

13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}

2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0

2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0

There are two solutions:

v_{2,f} = 0\,\frac{m}{s} or v_{2,f} = 3.606\,\frac{m}{s}

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: (v_{2,f} = 3.606\,\frac{m}{s})

v_{1,f} = 5.86-2.25\cdot (3.606)

v_{1,f} = -2.254\,\frac{m}{s}

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

3 0
3 years ago
Listed below are the measured radiation absorption rates​ (in W/kg) corresponding to 11 cell phones. Use the given data to const
Fantom [35]

Answer:

The 5-number summary is

1. Median = 0.93 W/kg

2. Lower quartile = 0.69 W/kg

3. Upper quartile = 1.16 W/kg

4. Minimum value = 0.54 W/kg

5. Maximum value = 1.42 W/kg

Explanation:

We are given the measured radiation absorption rates​ (in W/kg) corresponding to 11 cell phones.

1.16 0.85 0.69 0.75 0.95 0.93 1.18 1.17 1.42 0.54 0.57

What is 5-number summary?

A 5-number summary refers to a box plot that basically shows 5 statistical characteristics of a data set.

These statistical characteristics are:    

1. Median

2. Lower quartile

3. Upper quartile  

4. Minimum value  

5. Maximum value  

1. Median:

Arrange the data in ascending order

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

(n+1)/2 gives the median value of the data set.

(11 + 1)/2 = 6th position

Therefore, 0.93 W/kg is the median of the data set.

2. Lower quartile:

Divide the data set into two equal halfs (include median in both if n = odd)

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The lower quartile is the median of the lower half of the data set.

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

The median is 6/2 = 3rd position

Therefore, the lower quartile of the data set is 0.69 W/kg

3. Upper quartile:

Divide the data set into two equal halfs (include median in both if n = odd)

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The upper quartile is the median of the lower half of the data set.

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The median is 6/2 = 3rd position

Therefore, the upper quartile of the data set is 1.16 W/kg

4. Minimum value:

The minimum value is the least value in the data set.

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

Therefore, the minimum value of the data set is 0.54 W/kg

5. Maximum value  

The maximum value is the least value in the data set.

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

Therefore, the maximum value of the data set is 1.42 W/kg

The box plot is illustrated in the attached diagram.

6 0
3 years ago
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