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Mkey [24]
3 years ago
5

A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric

field zero?
Engineering
1 answer:
diamong [38]3 years ago
3 0

Answer:

Explanation:

Given

q_1=11.5\ nC charge is placed at x=0\ cm

another charge of q_2=-1.2\ nC is at x=3\ cm

We know that Electric field due to positive charge is away  from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to q_2 at some distance r from it

E_2=\frac{kq_2}{r^2}

Now Electric Field due to q_1 is

E_1=\frac{kq_1}{(3+r)^2}

Now E_1+E_2=0

\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0

\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}

\frac{3+r}{r}=3.095

thus r=1.43\ cm

Thus Electric field is zero at some distance r=1.43 cm right of q_2

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A long rod of 60-mm diameter and thermophysical properties rho= 8000 kg/m3, c= 500 J/kg·K, and k= 50 W/m·K is initially at a uni
Dvinal [7]

Answer:

Tc =    = 424.85 K

Explanation:

Data given:

D = 60 mm = 0.06 m

\rho = 8000 kg/m^3

k = 50 w/m . k

c = 500 j/kg.k

h_{\infty} = 1000 w/m^2

t_{\infity} = 750 k

t_w = 500 K

surface area = As = \pi dL

\frac{As}{L} = \pi D = \pi \timeS 0.06

HEAT FLOW Q  is

Q = h_{\infty} As (T_[\infty} - Tw)

 = 1000 \pi\times 0.06 (750-500)

  = 47123.88 w per unit length of rod

volumetric heat rate

q = \frac{Q}{LAs}

  = \frac{47123.88}{\frac{\pi}{4} D^2 \times 1}

q = 1.66\times 10^{7} w/m^3

Tc = \frac{- qR^2}{4K} + Tw

= \frac{ - 1.67\times 10^7 \times (\frac{0.06}{2})^2}{4\times 56} +  500

   = 424.85 K

7 0
3 years ago
The volume of 1.5 kg of helium in a frictionless piston-cylinder device is initially 6 m3. Now, helium is compressed to 2 m3 whi
coldgirl [10]

Answer:

The initial temperature will be "385.1°K" as well as final will be "128.3°K".

Explanation:

The given values are:

Helium's initial volume, v₁ = 6 m³

Mass, m = 1.5 kg

Final volume, v₂ = 2 m³

Pressure, P = 200 kPa

As we know,

Work, W=p(v_{2}-v_{1})

On putting the estimated values, we get

⇒            =200000(2-6)

⇒            =200000\times (-4)

⇒            =800,000 \ N.m

Now,

Gas ideal equation will be:

⇒  pv_{1}=mRT_{1}

On putting the values. we get

⇒  200000\times 6=1.5\times 2077\times T_{1}

⇒  T_{1}=\frac{1200000}{3115.5}

⇒       =385.1^{\circ}K (Initial temperature of helium)

and,

⇒  pv_{2}=mRT_{2}

On putting the values, we get

⇒  200000\times 2=1.5\times 2077\times T_{2}

⇒  T_{2}=\frac{400000}{3115.5}

⇒       =128.3^{\circ}K (Final temperature of helium)

3 0
3 years ago
Tech A says that the stoichiometric ratio is 14.7 gallons of air to 1 gallon of fuel. Tech B says it is 14.7 lb of air to 1 lb o
Degger [83]

The person that is correct as regards statement made about stoichiometric ratio is; <em><u>Tech B</u></em>

In air-fuel mixture, when there is a mix air and fuel in the ideal amounts to make an engine to run efficiently,the ideal ratio of air to fuel is known Stoichiometric ratio.

This stoichiometric ratio is the comparison between the weight of the air relative to the weight of the fuel that it has been mixed with.

Now, weight is in lb and as such, tech B is correct because his ratio is 14.7lb to 1 lb.

Read more at; brainly.com/question/16971089

3 0
2 years ago
What Is the Job Demand for Solar Engineers?<br> (2 or more sentences please and thank you)
podryga [215]

Answer:

repair of inverters and anything that has to do with solar energy...that is power from the sun

4 0
3 years ago
Water flows at a rate of 0.040 m3 /s in a horizontal pipe whose diameter is reduced from 15 cm to 8 cm by a reducer. If the pres
Blababa [14]

Answer:

hL = 0.9627 m

Explanation:

Given

Q = 0.040 m³/s (constant value)

D₁ = 15 cm = 0.15 m  ⇒  R₁ = D₁/2 = 0.15 m/2 = 0.075 m

D₂ = 8 cm = 0.08 m  ⇒  R₂ = D₂/2 = 0.08 m/2 = 0.04 m

P₁ = 480 kPa = 480*10³Pa

P₂ = 440 kPa = 440*10³Pa

α = 1.05

ρ = 1000 Kg/m³

g = 9.81 m/s²

h₁ = h₂

hL = ?  (the irreversible head loss in the reducer)

Using the formula Q = v*A   ⇒  v = Q/A

we can find the velocities v₁ and v₂ as follows

v₁ = Q/A₁ = Q/(π*R₁²) = (0.040 m³/s)/(π*(0.075 m)²) = 2.2635 m/s

v₂ = Q/A₂ = Q/(π*R₂²) = (0.040 m³/s)/(π*(0.04 m)²) = 7.9577 m/s

Then we apply the Bernoulli law (for an incompressible flow)

(P₂/(ρ*g)) + (α*v₂²/(2*g)) + h₂ = (P₁/(ρ*g)) + (α*v₁²/(2*g)) + h₁ - hL

Since h₁ = h₂ we obtain

(P₂/(ρ*g)) + (α*v₂²/(2*g)) = (P₁/(ρ*g)) + (α*v₁²/(2*g)) - hL

⇒  hL = ((P₁-P₂)/(ρ*g)) + (α/(2*g))*(v₁²-v₂²)

⇒  hL = ((480*10³Pa-440*10³Pa)/(1000 Kg/m³*9.81 m/s²)) + (1.05/(2*9.81 m/s²))*((2.2635 m/s)²-(7.9577 m/s)²)

⇒  hL = 0.9627 m

7 0
3 years ago
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