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3 years ago

8

The type of model that is a complete and unambiguous mathematical representation of a precisely enclosed and filled volume is a(

n) __________ model?

1 answer:

gayaneshka [121]3 years ago

5 0

A digital solid is a 3D model consisting of vertices, edges, faces, and partially filled or entirely filled interior. It is a complete and unambiguous representation of the object in a precisely enclosed and filled volume in digital space

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The mass flow rate in a 4.0-m wide, 2.0-m deep channel is 4000 kg/s of water. If the velocity distribution in the channel is lin

IceJOKER [234]

Answer:

V = 0.5 m/s

Explanation:

given data:

width of channel = 4 m

depth of channel = 2 m

mass flow rate = 4000 kg/s = 4 m3/s

we know that mass flow rate is given as

$\dot{m}=\rho AV$

Putting all the value to get the velocity of the flow

$\frac{\dot{m}}{\rho A} = V$

$V = \frac{4000}{1000*4*2}$

V = 0.5 m/s

4 0

3 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

Using the celsius_to_kelvin function as a guide, create a new function, changing the name to kelvin_to_celsius, and modifying th

aleksandr82 [10.1K]

Answer:

# kelvin_to_celsius function is defined

# it has value_kelvin as argument

def kelvin_to_celsius(value_kelvin):

# value_celsius is initialized to 0.0

value_celsius = 0.0

# value_celsius is calculated by

# subtracting 273.15 from value_kelvin

value_celsius = value_kelvin - 273.15

# value_celsius is returned

return value_celsius

# celsius_to_kelvin function is defined

# it has value_celsius as argument

def celsius_to_kelvin(value_celsius):

# value_kelvin is initialized to 0.0

value_kelvin = 0.0

# value_kelvin is calculated by

# adding 273.15 to value_celsius

value_kelvin = value_celsius + 273.15

# value_kelvin is returned

return value_kelvin

value_c = 0.0

value_k = 0.0

value_c = 10.0

# value_c = 10.0 is used to test the function celsius_to_kelvin

# the result is displayed

print(value_c, 'C is', celsius_to_kelvin(value_c), 'K')

value_k = 283.15

# value_k = 283.15 is used to test the function kelvin_to_celsius

# the result is displayed

print(value_k, 'is', kelvin_to_celsius(value_k), 'C')

Explanation:

Image of celsius_to_kelvin function used as guideline is attached

Image of program output is attached.

4 0

3 years ago

vertical gate in an irrigation canal holds back 12.2 m of water. Find the average force on the gate if its width is 3.60 m. Repo

DanielleElmas [232]

Answer:

The right solution is "2625 kN".

Explanation:

According to the question,

The average pressure will be:

= $density\times g\times \frac{h}{2}$

By putting values, we get

= $1000\times 9.8\times \frac{12.2}{2}$

= $1000\times 9.8\times 6.1$

= $59780$

hence,

The average force will be:

= $Pressure\times Area$

= $59780\times 3.6\times 12.2$

= $2625537 \ N$

Or,

= $2625 \ kN$

5 0

2 years ago

Whats the best used for cable -stayed bridge

nalin [4]

Answer:

a cable -stayed bridge has, one or more towers,from which cable support the bridge deck.

7 0

3 years ago