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laiz [17]
3 years ago
9

A charge of 11.748 nC is uniformly distributed along the x-axis from −2 m to 2 m . What is the electric potential (relative to z

ero at infinity) of the point at 5 m on the x-axis? The value of the Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of V
Engineering
1 answer:
svp [43]3 years ago
6 0

Answer:

electric potential  =  22.36 volt

Explanation:

given data

charge Q =  11.748 nC

distance d = 5 - 2 = 3 m

length = 2 + 2 = 4 m

Coulomb constant = 8.98755 × 109 N·m²/C ²

solution

electric potential is express as

electric potential  = \frac{Q}{4\pi \epsilon  _o L} ln(1+\frac{L}{d})   ..............1

electric potential  = \frac{KQ}{L} ln(1+\frac{L}{d})  

put here value

electric potential  = \frac{8.98755\times 10^9\times 11.748\times 10{-9}}{4} ln(1+\frac{4}{3})  

electric potential  =  22.36 volt

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Steam enters an insulated turbine at 100 bar, 400oC. At the exit, the pressure and quality are 200 kPa and 0.47, respectively. D
OverLord2011 [107]

Answer:

power produced = 3098.52 kW

Explanation:

given data

insulated turbine = 100 bar

temperature = 400°C

pressure = 200 kPa

mass flow rate = 1.99 kg/s

solution

we use here steam table for At 100 bar and 400°C  

h1 = 3096.5 KJ/Kg

and  

at P2 = 200 Kpa

h2 = hf + 0.47 hg

h2 = 504.7 + 0.47 ×  2201.6  

h2 = 1539.452 KJ/Kg

so here

power produced  is express as

power produced = m × (h1 - h2)    .................1

power produced = 1.99 × ( 3096.5 - 1539.452 )

power produced = 3098.52 kW

8 0
4 years ago
A very large thin plate is centered in a gap of width 0.06 m with a different oils of unknown viscosities above and below; one v
In-s [12.5K]

Answer:

The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s

Explanation:

Assuming the two oils are Newtonian fluids.

From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.

τ = μ (∂v/∂y)

There are oils above and below the plate, so we can write this expression for the both cases.

τ₁ = μ₁ (∂v/∂y)

τ₂ = μ₂ (∂v/∂y)

dv = 0.3 m/s

dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)

τ₁ = μ₁ (0.3/0.03) = 10μ₁

τ₂ = μ₂ (0.3/0.03) = 10μ₂

But the shear stress on the plate is given as 29 N per square meter.

τ = 29 N/m²

But this stress is a sum of stress due to both shear stress above and below the plate

τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29

But it is also given that one viscosity is twice the other

μ₁ = 2μ₂

10μ₁ + 10μ₂ = 29

10(2μ₂) + 10μ₂ = 29

30μ₂ = 29

μ₂ = (29/30) = 0.967 Pa.s

μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s

Hope this Helps!!!

6 0
3 years ago
Find R subscript C and R subscript B in the following circuit such that BJT would be in the active region with V subscript C E e
Alex777 [14]

Answer: Rc = 400 Ω and Rb = 57.2 kΩ

Explanation:

Given that;

VCE = 5V

VCC = 15 V

iC = 25 mA

β = 100

VD₀ = 0.7 V

taking a look at the image; at loop 1

-VCC + (i × Rc) + VCE = 0

we substitute

-15 + ( 25 × Rc) + 5 = 0

25Rc = 10

Rc = 10 / 25

Rc = 0.4 k

Rc = 0.4 × 1000

Rc = 400 Ω

iC = βib

25mA = 100(ib)

ib = 25 mA / 100

ib = 0.25 mA

ib = 0.25 × 1000

ib = 250 μAmp

Now at Loop 2

-Vcc + (ib×Rb) + VD₀ = 0

-15 (250 × Rb) + 0.7 = 0

250Rb = 15 - 0.7

250Rb = 14.3

Rb = 14.3 / 250

Rb = 0.0572 μ

Rb = 0.0572 × 1000

Rb = 57.2 kΩ

Therefore Rc = 400 Ω and Rb = 57.2 kΩ

8 0
3 years ago
You have a 12 volt power source running through a circuit that has 3kΩ of resistance, how many amps (in mA) can flow through the
siniylev [52]

Answer:

4mA

Explanation:

For this problem, we will simply apply Ohm's law:

V = IR

V/R = I

I = V / R

I = 12 volt / 3kΩ

I = 4mA

Hence, the current in the circuit is 4mA.

Cheers.

5 0
3 years ago
Are connected to a larger parent organization that can help with marketing and provide will have the same
Setler [38]
Answer: D. Franchises

7 0
3 years ago
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