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Flura [38]
3 years ago
15

Particles q1, q2, and q3 are in a straight line. Particles q1 = -28.1 μC, q2 = +25.5 μC, and q3 = -47.9 μC. Particles q1 and q2

are separated by 0.300 m. Particles q2 and q3 are separated by 0.300 m. What is the net force on q3?
Physics
2 answers:
Arturiano [62]3 years ago
3 0

Answer:-88.4

Explanation:

Digiron [165]3 years ago
3 0

Answer:

-88.4

Explanation:

hope it helps

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Feliz [49]
It is absorbed, or can be reflected by clouds, gasses, dust. or is reflected off of Earths surface.
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3 years ago
Una tubería con secciones circulares se emplea para el traslado de agua entre el tanque central y un tanque auxiliar. El tanque
charle [14.2K]

A) 20 L/s

B) 2.55 m/s

C) 10.20 m/s

D) 400.8 kPa

Explanation:

a)

In this problem, we know that the volume of the tank:

V=72 m^3  

is filled in a time of

t = 1 h

We also know that the volume flow rate of water in the pipe is constant in every point of the pipe, so it can be calculated directly from the two data above.

The volume of the tank, in liter, is

V=72 m^3 = 72,000 L

While the time, in seconds, is

t=1 h = 3600 s

Therefore, the volume flow rate in Liters per second is:

Q=\frac{V}{t}=\frac{72,000}{3600}=20 L/s

b)

The volume flow rate of water through the pipe can be also written as

Q=Av

where

A is the cross-sectional area of the pipe

v is the speed of the water

In section B, we have:

r = 0.050 m is the radius of section B

so, the cross-sectional area of section B is:

A=\pi r^2 = \pi (0.050)^2=0.00785 m^2

The volume flow rate in SI units is

Q=\frac{V}{t}=\frac{72.0m^3}{3600 s}=0.02 m^3/s

Therefore, the speed of the water in section B is:

v=\frac{Q}{A}=\frac{0.02}{0.00785}=2.55 m/s

c)

As in part B), we know that the volume flow rate must remain constant through the entire pipe.

So, the volume flow rate in section A of the pipe is still

Q=0.02 m^3/s

The radius of the pipe in section A is

r=0.025 m

Therefore, the cross-sectional area in section A of the pipe is

A=\pi r^2 = \pi (0.025)^2=0.00196 m^2

So, since we have

Q=Av

we can find the speed of water in section A:

v=\frac{Q}{A}=\frac{0.02}{0.00196}=10.20 m/s

d)

Here we want to find the gauge pressure in section B.

We know that:

p_A = 2.0 atm = 105,000 Pa is the pressure in section A

h_A=15.0m is the altitude of section A

v_A=10.20 m/s is the speed of water in section A

v_B=2.55 m/s is the speed of water in section B

We can write Bernoulli's equation:

p_A + \rho g h_A + \frac{1}{2}\rho v_A^2 = p_B + \frac{1}{2}\rho v_B^2

where

\rho=1000 kg/m^3 is the water density

p_B is the pressure in section B

And solving for pB, we find:

p_B = p_A + \rho g h_A + \frac{1}{2}\rho (v_A^2-v_B^2)=\\=205,000+(1000)(9.8)(15.0)+\frac{1}{2}(1000)(10.20^2-2.55^2)=400,769 Pa

Which is

p_B = 400.8 kPa

5 0
3 years ago
PLEASE HELP!!!!
OLEGan [10]

The story that fits the graph is; he elevator moves up, is stationary for a while, then moves down at a slower rate.

<h3>What is a position time graph?</h3>

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We now have to turn to the graph and see how the graph can be able to give a description of the motion of the elevator. If we look at the position time graph, it tends to move up and then remain at a point for some time before it moves down.

Based on the analysis that we have made above, we can now look out for any of the stories that fits the description that we have just made here.

Learn more about position time graph:brainly.com/question/13693087

#SPJ1

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I’m which medium does sound travel fastest railroad track or across the room
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The closer the particles, the more will be the propogation of sound waves. Room contains air molecules which are far away from each other. So it takes much time for one molecule of air to disturb the other one. But in case of solids, as particles are much closer(compared to fluids), disturbance generated by one molecule is quickly transmitted to the next molecule
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What happens to the atom if one proton and one electron is removed from the atom?
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In that case, it will change it's identity to the atom of one degree less than that!

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