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iris [78.8K]
2 years ago
10

The net electric flux through an octahedron is -1850 N⋅m2/C . Part A How much charge is enclosed within the octahedron? Express

your answer with the appropriate units.
Physics
1 answer:
Masteriza [31]2 years ago
8 0

The electric flow is defined as the change of the charge in relation to the constant of the vacuum permittivity constant.

That is to say that mathematically this expression could be given as

\Phi = \frac{q}{\epsilon_0}

Here,

\Phi = Electric Flux

q = Charge

\epsilon_0 = Vacuum Permittivity

Replacing with our values and solving to find q, we have,

-1850N\cdot m^2/C = \frac{q}{8.854*10^{-12} C\cdot N^{-1} \cdot m^{-2}}

q = - 1.63799*10^{-8}C

Therefore the charge is enclose within the octahedron with a charge of - 1.63799*10^{-8}C

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Alicia can row 6 miles downstream in the same time it takes her to row 4 miles upstream. She rows downstream 3 miles/hour faster
m_a_m_a [10]
Let us assume the upstream rowing rate of Alicia = x
Let us assume the downstream rowing rate of Alicia = y
We already know that
Travelling time = Distance traveled/rowing rate
Then
6/(x + 3) = 4/x
6x = 4x + 12
6x - 4x = 12
2x = 12
x = 6
Then
Rowing rate of Alicia going upstream = 6 miles per hour
Rowing rate of Alicia going downstream = 9 miles per hour.
4 0
2 years ago
Read 2 more answers
A heat engine (Power Cycle) with a thermal efficiency of 35 percent efficiency produces 750 kJ of work. Heat transfer to the eng
frosja888 [35]

Answer:

a) The schematic illustrating is attached

b) The heat transfer to the heat engine is 2142.86 kJ, the heat transfer from the heat engine is 1392.86 kJ

c) The heat transfer to the heat engine is 1648.35 kJ, the heat transfer from the heat engine is 898.35 kJ

Explanation:

b) The heat transfer to the engine and the heat transfer from the engine to the air is:

Q_{1} =\frac{W}{n}

Where

W = 750 kJ

n = 35% = 0.25

Replacing:

Q_{1} =\frac{750}{0.35} =2142.86kJ

Q_{2} =Q_{1} -W=2142.86-750=1392.86kJ

c) The efficiency of Carnot engine is:

n=1-\frac{300K}{550K} =0.455

The heat transfer to the heat engine is:

Q_{1c} =\frac{750}{0.455} =1648.35kJ

The heat transfer from the heat engine is:

Q_{2c} =1648.35-750=898.35kJ

4 0
2 years ago
In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad
Crank

Answer:

a)   Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c)      Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

        P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

         A₁ v₁ = A₂ v₂

         v₂ = v₁ A₁ / A₂

We replace

         ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

         ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

         v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

           Q = A v

           Q = A₁ v₁

           Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]

The areas are

            A₁ = π r₁

            A₂ = π r₂

We replace

        Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]

Let's calculate for the different pressures

      r₁ = d₁ / 2 = 1.00 / 2

      r₁ = 0.500 10⁻² m

      r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

      Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))

       Q = 1.57 10⁻² √(12 10³/2550)

        Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

        Q = 1.57 10⁻² √(2 12 10³ / (850 3)

         Q = 4.8 10⁻² m³ / s

5 0
2 years ago
A straight wire lies along the y-axis initially carrying a current of 10 A in the positive y-direction. The current decreases an
Elan Coil [88]

Answer:

Explanation:

The magnetic field due to straight wire is into the square coil.

As the current in straight wire decreases the magnetic flux in the coil decreases . The induced magnetic field is into the coil.The induced current is along +y direction

8 0
2 years ago
A cylindrical tank of methanol has a mass of 40 kgand a volume of 51 L. Determine the methanol’s weight, density,and specific gr
mezya [45]

Answer:

Weight  W = 392.4 N

Density  \rho = 784.31 \frac{kg}{m^{3} }

Specific gravity S = 0.78431

Force required F = 10 N

Explanation:

Given data

Mass (m) = 40 kg

Volume (V) = 0.051 m^{3}

Weight W = m × g

⇒ W = 40 × 9.81

⇒ W = 392.4 N

This is the weight of the methanol.

Density \rho = \frac{mass }{volume}

⇒ \rho = \frac{40}{0.051}

⇒ \rho = 784.31 \frac{kg}{m^{3} }

This is the density of the methanol.

Specific gravity (S) = \frac{\rho}{\rho_{water} }

⇒ S = \frac{784.31}{1000}

⇒ S = 0.78431

This is the specific gravity of the methanol.

Force needed to accelerate this tank F = ma

⇒ F = 40 × 0.25

⇒ F = 10 N

This is the force required to accelerate the tank.

4 0
2 years ago
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